2019的hdu暑假作业（欢迎纠错）

6月26日

被1226烦得头都大了，从下午断断续续调到晚。最后数组开小给报个Tle看了半天没看出来...

6月30日

每日补题，进度停滞中...

7月10日

第一阶段结束...不做鸽子了不做鸽子了。

7月11日

被卡了....还没题解....生命--。

7月21日

达到40题之后已经咸鱼了。随缘做做。

hdu1219

遍历计数。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 char a[100010];
 7 int cnt[200];
 8 int main(){
 9     while(cin.getline(a,100005)){
10         memset(cnt,0,sizeof(cnt));
11         int l=strlen(a);
12         for(int i=0;i<l;i++){
13             cnt[a[i]]++;
14         }
15         for(int i='a';i<='z';i++){
16             printf("%c:%d\n",i,cnt[i]);
17         }
18         printf("\n");
19     }
20     return QAQ;
21 }

hdu 1219

hdu1220

总对数减去相邻的对数。相邻的对数分类为顶点、棱、面、体内的小正方体，分别对应邻接3、4、5、6个，每对算了两次。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 typedef long long ll;
 6 
 7 ll n;
 8 
 9 int main(){
10     while(scanf("%lld",&n)!=EOF){
11         if(n==1){
12             printf("%lld\n",0);
13             continue;
14         }
15         ll sum=n*n*n*(n*n*n-1)/2,cnt;
16         cnt=8*3 + 12*(n-2)*4 + 6*(n-2)*(n-2)*5 + (n-2)*(n-2)*(n-2)*6;
17         printf("%lld\n",sum-cnt/2);
18     }
19     return QAQ;
20 }

hdu 1220

hdu1221

最大值一定到在圆心到矩形四个顶点里，最小值除了到顶点距离外，再判一下圆心到四条边距离。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 typedef long long ll;
 6 const double eps=1e-7;
 7 
 8 double x,y,r,a1,b1,a2,b2,a3,b3,a4,b4,maxi,mini;
 9 ll T;
10 
11 double dis(double x1, double y1, double x2, double y2){
12     return sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
13 }
14 
15 int main(){
16     scanf("%lld",&T);
17     while(T--){
18         scanf("%lf%lf%lf%lf%lf%lf%lf",&x,&y,&r,&a1,&b1,&a2,&b2);
19         a3=a1; b3=b2; a4=a2; b4=b1;
20         double l1=dis(x,y,a1,b1),l2=dis(x,y,a2,b2),
21         l3=dis(x,y,a3,b3),l4=dis(x,y,a4,b4);
22         
23         maxi=max( max(l1,l2) , max(l3,l4) );
24         mini=min( min(l1,l2) , min(l3,l4) );
25         
26         if(x>a1 && x<a2){
27             mini=min( mini, min( fabs(y-b1) , fabs(y-b2) ) );
28         }
29 
30         if(y>b1 && y<b2){
31             mini=min( mini, min( fabs(x-a1) , fabs(x-a2) ) );
32         }
33 
34         if(mini<=r && maxi>=r){
35             printf("YES");
36         }
37         else printf("NO");
38         printf("\n");
39     }
40     return QAQ;
41 }

hdu 1221

hdu1222

如果m和n的最大公约数不为1，则一定有安全位置。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int T,m,n;
 7 
 8 int gcd(int a,int b){
 9     return b==0?a:gcd(b,a%b);
10 }
11 
12 int main(){
13     scanf("%d",&T);
14     while(T--){
15         scanf("%d%d",&m,&n);
16         if(gcd(m,n)!=1) printf("YES");
17         else printf("NO");
18         printf("\n");
19     }
20     return QAQ;
21 }

hdu 1222

hdu1223

有等号连接的为一个块，dp[i][j]表示i个字母有j个块。对于i-1个字母分两类：1.原有j块，即添加等号，显然和每个块相等为一种，共j种；2.原有j-1块，即添加小于号，j-1个块有j个空位可以放入，也为j种。所以dp[i][j]=dp[i-1][j]*j+dp[i-1]][j-1]*j。

本题没有mod，要用大数。感谢颜神的模板支援o(*￣▽￣*)ブ。

  1 #include<bits/stdc++.h>
  2 #define QAQ 0
  3 
  4 using namespace std;
  5 typedef long long ll;
  6 
  7 int T,n;
  8 
  9 const int base = 1000000000;
 10 const int base_digits = 9;
 11 
 12 struct Bigint {
 13     vector<int> z;
 14     int sign;
 15 
 16     Bigint() :
 17         sign(1) {
 18     }
 19 
 20     Bigint(long long v) {
 21         *this = v;
 22     }
 23 
 24     Bigint(const string &s) {
 25         read(s);
 26     }
 27 
 28     void operator=(const Bigint &v) {
 29         sign = v.sign;
 30         z = v.z;
 31     }
 32 
 33     void operator=(long long v) {
 34         sign = 1;
 35         if (v < 0)
 36             sign = -1, v = -v;
 37         z.clear();
 38         for (; v > 0; v = v / base)
 39             z.push_back(v % base);
 40     }
 41 
 42     Bigint operator+(const Bigint &v) const {
 43         if (sign == v.sign) {
 44             Bigint res = v;
 45 
 46             for (int i = 0, carry = 0; i < (int) max(z.size(), v.z.size()) || carry; ++i) {
 47                 if (i == (int) res.z.size())
 48                     res.z.push_back(0);
 49                 res.z[i] += carry + (i < (int) z.size() ? z[i] : 0);
 50                 carry = res.z[i] >= base;
 51                 if (carry)
 52                     res.z[i] -= base;
 53             }
 54             return res;
 55         }
 56         return *this - (-v);
 57     }
 58 
 59     Bigint operator-(const Bigint &v) const {
 60         if (sign == v.sign) {
 61             if (abs() >= v.abs()) {
 62                 Bigint res = *this;
 63                 for (int i = 0, carry = 0; i < (int) v.z.size() || carry; ++i) {
 64                     res.z[i] -= carry + (i < (int) v.z.size() ? v.z[i] : 0);
 65                     carry = res.z[i] < 0;
 66                     if (carry)
 67                         res.z[i] += base;
 68                 }
 69                 res.trim();
 70                 return res;
 71             }
 72             return -(v - *this);
 73         }
 74         return *this + (-v);
 75     }
 76 
 77     void operator*=(int v) {
 78         if (v < 0)
 79             sign = -sign, v = -v;
 80         for (int i = 0, carry = 0; i < (int) z.size() || carry; ++i) {
 81             if (i == (int) z.size())
 82                 z.push_back(0);
 83             long long cur = z[i] * (long long) v + carry;
 84             carry = (int) (cur / base);
 85             z[i] = (int) (cur % base);
 86             //asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));
 87         }
 88         trim();
 89     }
 90 
 91     Bigint operator*(int v) const {
 92         Bigint res = *this;
 93         res *= v;
 94         return res;
 95     }
 96 
 97     friend pair<Bigint, Bigint> divmod(const Bigint &a1, const Bigint &b1) {
 98         int norm = base / (b1.z.back() + 1);
 99         Bigint a = a1.abs() * norm;
100         Bigint b = b1.abs() * norm;
101         Bigint q, r;
102         q.z.resize(a.z.size());
103 
104         for (int i = a.z.size() - 1; i >= 0; i--) {
105             r *= base;
106             r += a.z[i];
107             int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
108             int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
109             int d = ((long long) s1 * base + s2) / b.z.back();
110             r -= b * d;
111             while (r < 0)
112                 r += b, --d;
113             q.z[i] = d;
114         }
115 
116         q.sign = a1.sign * b1.sign;
117         r.sign = a1.sign;
118         q.trim();
119         r.trim();
120         return make_pair(q, r / norm);
121     }
122 
123     friend Bigint sqrt(const Bigint &a1) {
124         Bigint a = a1;
125         while (a.z.empty() || a.z.size() % 2 == 1)
126             a.z.push_back(0);
127 
128         int n = a.z.size();
129 
130         int firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]);
131         int norm = base / (firstDigit + 1);
132         a *= norm;
133         a *= norm;
134         while (a.z.empty() || a.z.size() % 2 == 1)
135             a.z.push_back(0);
136 
137         Bigint r = (long long) a.z[n - 1] * base + a.z[n - 2];
138         firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]);
139         int q = firstDigit;
140         Bigint res;
141 
142         for(int j = n / 2 - 1; j >= 0; j--) {
143             for(; ; --q) {
144                 Bigint r1 = (r - (res * 2 * base + q) * q) * base * base + (j > 0 ? (long long) a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0);
145                 if (r1 >= 0) {
146                     r = r1;
147                     break;
148                 }
149             }
150             res *= base;
151             res += q;
152 
153             if (j > 0) {
154                 int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
155                 int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
156                 int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
157                 q = ((long long) d1 * base * base + (long long) d2 * base + d3) / (firstDigit * 2);
158             }
159         }
160 
161         res.trim();
162         return res / norm;
163     }
164 
165     Bigint operator/(const Bigint &v) const {
166         return divmod(*this, v).first;
167     }
168 
169     Bigint operator%(const Bigint &v) const {
170         return divmod(*this, v).second;
171     }
172 
173     void operator/=(int v) {
174         if (v < 0)
175             sign = -sign, v = -v;
176         for (int i = (int) z.size() - 1, rem = 0; i >= 0; --i) {
177             long long cur = z[i] + rem * (long long) base;
178             z[i] = (int) (cur / v);
179             rem = (int) (cur % v);
180         }
181         trim();
182     }
183 
184     Bigint operator/(int v) const {
185         Bigint res = *this;
186         res /= v;
187         return res;
188     }
189 
190     int operator%(int v) const {
191         if (v < 0)
192             v = -v;
193         int m = 0;
194         for (int i = z.size() - 1; i >= 0; --i)
195             m = (z[i] + m * (long long) base) % v;
196         return m * sign;
197     }
198 
199     void operator+=(const Bigint &v) {
200         *this = *this + v;
201     }
202     void operator-=(const Bigint &v) {
203         *this = *this - v;
204     }
205     void operator*=(const Bigint &v) {
206         *this = *this * v;
207     }
208     void operator/=(const Bigint &v) {
209         *this = *this / v;
210     }
211 
212     bool operator<(const Bigint &v) const {
213         if (sign != v.sign)
214             return sign < v.sign;
215         if (z.size() != v.z.size())
216             return z.size() * sign < v.z.size() * v.sign;
217         for (int i = z.size() - 1; i >= 0; i--)
218             if (z[i] != v.z[i])
219                 return z[i] * sign < v.z[i] * sign;
220         return false;
221     }
222 
223     bool operator>(const Bigint &v) const {
224         return v < *this;
225     }
226     bool operator<=(const Bigint &v) const {
227         return !(v < *this);
228     }
229     bool operator>=(const Bigint &v) const {
230         return !(*this < v);
231     }
232     bool operator==(const Bigint &v) const {
233         return !(*this < v) && !(v < *this);
234     }
235     bool operator!=(const Bigint &v) const {
236         return *this < v || v < *this;
237     }
238 
239     void trim() {
240         while (!z.empty() && z.back() == 0)
241             z.pop_back();
242         if (z.empty())
243             sign = 1;
244     }
245 
246     bool isZero() const {
247         return z.empty() || (z.size() == 1 && !z[0]);
248     }
249 
250     Bigint operator-() const {
251         Bigint res = *this;
252         res.sign = -sign;
253         return res;
254     }
255 
256     Bigint abs() const {
257         Bigint res = *this;
258         res.sign *= res.sign;
259         return res;
260     }
261 
262     long long longValue() const {
263         long long res = 0;
264         for (int i = z.size() - 1; i >= 0; i--)
265             res = res * base + z[i];
266         return res * sign;
267     }
268 
269     friend Bigint gcd(const Bigint &a, const Bigint &b) {
270         return b.isZero() ? a : gcd(b, a % b);
271     }
272     friend Bigint lcm(const Bigint &a, const Bigint &b) {
273         return a / gcd(a, b) * b;
274     }
275 
276     void read(const string &s) {
277         sign = 1;
278         z.clear();
279         int pos = 0;
280         while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) {
281             if (s[pos] == '-')
282                 sign = -sign;
283             ++pos;
284         }
285         for (int i = s.size() - 1; i >= pos; i -= base_digits) {
286             int x = 0;
287             for (int j = max(pos, i - base_digits + 1); j <= i; j++)
288                 x = x * 10 + s[j] - '0';
289             z.push_back(x);
290         }
291         trim();
292     }
293 
294     friend istream& operator>>(istream &stream, Bigint &v) {
295         string s;
296         stream >> s;
297         v.read(s);
298         return stream;
299     }
300 
301     friend ostream& operator<<(ostream &stream, const Bigint &v) {
302         if (v.sign == -1)
303             stream << '-';
304         stream << (v.z.empty() ? 0 : v.z.back());
305         for (int i = (int) v.z.size() - 2; i >= 0; --i)
306             stream << setw(base_digits) << setfill('0') << v.z[i];
307         return stream;
308     }
309 
310     static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
311         vector<long long> p(max(old_digits, new_digits) + 1);
312         p[0] = 1;
313         for (int i = 1; i < (int) p.size(); i++)
314             p[i] = p[i - 1] * 10;
315         vector<int> res;
316         long long cur = 0;
317         int cur_digits = 0;
318         for (int i = 0; i < (int) a.size(); i++) {
319             cur += a[i] * p[cur_digits];
320             cur_digits += old_digits;
321             while (cur_digits >= new_digits) {
322                 res.push_back(int(cur % p[new_digits]));
323                 cur /= p[new_digits];
324                 cur_digits -= new_digits;
325             }
326         }
327         res.push_back((int) cur);
328         while (!res.empty() && res.back() == 0)
329             res.pop_back();
330         return res;
331     }
332 
333     typedef vector<long long> vll;
334 
335     static vll karatsubaMultiply(const vll &a, const vll &b) {
336         int n = a.size();
337         vll res(n + n);
338         if (n <= 32) {
339             for (int i = 0; i < n; i++)
340                 for (int j = 0; j < n; j++)
341                     res[i + j] += a[i] * b[j];
342             return res;
343         }
344 
345         int k = n >> 1;
346         vll a1(a.begin(), a.begin() + k);
347         vll a2(a.begin() + k, a.end());
348         vll b1(b.begin(), b.begin() + k);
349         vll b2(b.begin() + k, b.end());
350 
351         vll a1b1 = karatsubaMultiply(a1, b1);
352         vll a2b2 = karatsubaMultiply(a2, b2);
353 
354         for (int i = 0; i < k; i++)
355             a2[i] += a1[i];
356         for (int i = 0; i < k; i++)
357             b2[i] += b1[i];
358 
359         vll r = karatsubaMultiply(a2, b2);
360         for (int i = 0; i < (int) a1b1.size(); i++)
361             r[i] -= a1b1[i];
362         for (int i = 0; i < (int) a2b2.size(); i++)
363             r[i] -= a2b2[i];
364 
365         for (int i = 0; i < (int) r.size(); i++)
366             res[i + k] += r[i];
367         for (int i = 0; i < (int) a1b1.size(); i++)
368             res[i] += a1b1[i];
369         for (int i = 0; i < (int) a2b2.size(); i++)
370             res[i + n] += a2b2[i];
371         return res;
372     }
373 
374     Bigint operator*(const Bigint &v) const {
375         vector<int> a6 = convert_base(this->z, base_digits, 6);
376         vector<int> b6 = convert_base(v.z, base_digits, 6);
377         vll a(a6.begin(), a6.end());
378         vll b(b6.begin(), b6.end());
379         while (a.size() < b.size())
380             a.push_back(0);
381         while (b.size() < a.size())
382             b.push_back(0);
383         while (a.size() & (a.size() - 1))
384             a.push_back(0), b.push_back(0);
385         vll c = karatsubaMultiply(a, b);
386         Bigint res;
387         res.sign = sign * v.sign;
388         for (int i = 0, carry = 0; i < (int) c.size(); i++) {
389             long long cur = c[i] + carry;
390             res.z.push_back((int) (cur % 1000000));
391             carry = (int) (cur / 1000000);
392         }
393         res.z = convert_base(res.z, 6, base_digits);
394         res.trim();
395         return res;
396     }
397 }dp[60][60],sum[60];
398 
399 int main(){
400     scanf("%d",&T);
401     for(int i=1;i<=50;i++){
402         dp[i][1]=1;
403     }
404     for(int i=2;i<=50;i++){
405         for(int j=2;j<=i;j++){
406             dp[i][j]=dp[i-1][j]*j + dp[i-1][j-1]*j;
407         }
408     }
409     for(int i=1;i<=50;i++){
410         for(int j=1;j<=i;j++){
411             sum[i]=sum[i]+dp[i][j];
412         }
413     }
414     while(T--){
415         scanf("%d",&n);
416         cout<<sum[n]<<'\n';
417     }
418     return QAQ;
419 }

hdu 1223

hdu1224

数据很小，dfs一遍，顺带记录一下走的路线。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int T,n,cas,maxi,m,u,v,a[102];
 7 vector<int>ve[102],tmp,ans;
 8 
 9 void dfs(int x,int t){
10     if(x==n+1){
11         if(t>maxi){
12             maxi=t;
13             ans.clear();
14             ans=tmp;
15         }
16         return;
17     }
18     for(auto y:ve[x]){
19         tmp.push_back(x);
20         dfs(y,t+a[y]);
21         tmp.pop_back();
22     }
23 }
24 
25 int main(){
26     scanf("%d",&T);
27     while(T--){
28         maxi=-1;
29         tmp.clear(); ans.clear();
30         scanf("%d",&n);
31         for(int i=1;i<=n;i++){
32             scanf("%d",&a[i]);
33             ve[i].clear();
34         }
35         a[n+1]=a[1];
36         scanf("%d",&m);
37         for(int i=1;i<=m;i++){
38             scanf("%d%d",&u,&v);
39             ve[u].push_back(v);
40         }
41         dfs(1,a[1]);
42         cas++;
43         printf("CASE %d#\npoints : %d\ncircuit : ",cas,maxi);
44         for(auto x:ans){
45             printf("%d->",x);
46         }
47         printf("1\n");
48         if(T) printf("\n");
49     }
50     return QAQ;
51 }

hdu 1224

hdu1225

按题意模拟，用map计数，排序输出。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 #define miaojie
 4 #ifdef miaojie
 5   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
 6 void err() {std::cout << std::endl;}
 7 template<typename T, typename...Args>
 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
 9 #else
10   #define dbg(...)
11 #endif
12 
13 using namespace std;
14 
15 struct node{
16     string ss;
17     int r,r2,r3;
18     bool operator <(const node a) const{
19         if(r==a.r){
20             if(r2==a.r2){
21                 if(r3==a.r3){
22                     return ss<a.ss;
23                 }
24                 return r3>a.r3;
25             }
26             return r2>a.r2;
27         }
28         return r>a.r;
29     }
30 };
31 
32 int n,a,b;
33 char c;
34 string s1,s2,s3;
35 map<string,int>m,m2,m3;
36 set<string>s;
37 
38 int main(){
39     while(cin>>n){
40         m.clear(); m2.clear(); m3.clear(); s.clear();
41         for(int i=1;i<=n*(n-1);i++){
42             cin>>s1>>s2>>s3;
43             cin>>a>>c>>b;
44             s.insert(s1); s.insert(s3);
45             m2[s1]=m2[s1]+a-b;
46             m2[s3]=m2[s3]+b-a;
47             m3[s1]+=a;
48             m3[s3]+=b;
49             if(a>b){
50                 m[s1]+=3;
51             }
52             else if(a==b){
53                 m[s1]++;
54                 m[s3]++;
55             }
56             else{
57                 m[s3]+=3;
58             }    
59         }
60         
61         node ans[n];
62         int p=0;
63         for(auto x:s){
64             ans[p].r=m[x];
65             ans[p].r2=m2[x];
66             ans[p].r3=m3[x];
67             ans[p].ss=x;
68             p++;
69         }
70         sort(ans,ans+p);
71         for(int i=0;i<p;i++){
72             cout<<ans[i].ss<<" "<<ans[i].r<<endl;
73         }
74         cout<<'\n';
75     }
76     return QAQ;
77 }

hdu 1225

hdu1226

题目给了10s...从高位到低位bfs，这样保证一定从小到大，直到搜到模n为0的。模n相同的数等价，所以前面出现过的余数后面再有就不必入队了。n=0特判。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 #define miaojie1
 4 #ifdef miaojie
 5   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
 6 void err() {std::cout << std::endl;}
 7 template<typename T, typename...Args>
 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
 9 #else
10   #define dbg(...)
11 #endif
12 
13 using namespace std;
14 
15 int T,n,c,m;
16 char ch[5];
17 bool f[20],r[5005];
18 
19 struct node{
20     int v[505],l,rr;
21 };
22 
23 void bfs(){
24     queue<node>q;
25     for(int i=1;i<=15;i++){
26         if(f[i]){
27             node t;
28             t.v[0]=i;
29             t.rr=i%n;
30             if(r[t.rr]) continue;
31             r[t.rr]=1;
32             t.l=1;
33             if(!t.rr){
34                 if(i>=10){
35                     printf("%c",i+55);
36                 }
37                 else printf("%d",i);
38                 return;
39             }
40             q.push(t);
41         }
42     }
43     while(!q.empty()){
44         node t=q.front();
45         q.pop();
46         for(int i=0;i<=15;i++){
47             if(f[i]){
48                 node tt=t;
49                 tt.rr=(tt.rr*c+i)%n;
50                 tt.v[tt.l++]=i;
51                 if(!tt.rr){
52                     for(int j=0;j<tt.l;j++){
53                         if(tt.v[j]>=10){
54                             printf("%c",tt.v[j]+55);
55                         }
56                         else printf("%d",tt.v[j]);
57                     }
58                     dbg(tt.l);
59                     return;
60                 }
61                 if(!r[tt.rr] && tt.l<=499){
62                     r[tt.rr]=1;
63                     q.push(tt);
64                 }
65             }
66         }
67     }
68     printf("give me the bomb please");
69 }
70 
71 int main(){
72     scanf("%d",&T);
73     while(T--){
74         scanf("%d%d%d",&n,&c,&m);
75         memset(f,0,sizeof(f));
76         memset(r,0,sizeof(r));
77         
78         for(int i=1;i<=m;i++){
79             scanf("%s",ch);
80             if(ch[0]>='A'){
81                 f[ch[0]-55]=1;
82             }
83             else f[ch[0]-48]=1;
84         }
85         if(n==0){
86             if(f[0]){
87                 printf("0\n");
88             }
89             else printf("give me the bomb please\n");
90             continue;
91         }
92         bfs();
93         printf("\n");
94     }
95     return QAQ;
96 }

hdu 1226

hdu 1256

根据下面内圈一定是正方形推一下各个长度。题目表达的意思是竖线宽为n/6+1....读题读错了。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int T,n,a,b,d;
 7 char c,s[4];
 8 
 9 void f(){
10     for(int i=1;i<=d;i++){
11         printf(" ");
12     }
13     for(int i=1;i<=b;i++){
14         printf("%c",c);
15     }
16     printf("\n");
17 }
18 
19 void f2(int x){
20     for(int i=1;i<=x;i++){
21         for(int j=1;j<=d;j++){
22             printf("%c",c);
23         }
24         for(int j=1;j<=b;j++){
25             printf(" ");
26         }
27         for(int j=1;j<=d;j++){
28             printf("%c",c);
29         }
30         printf("\n");
31     }
32 }
33 
34 int main(){
35     scanf("%d",&T);
36     while(T--){
37         scanf("%s%d",s,&n);
38         c=s[0];
39         a=(n-3)/2,b=(n-2)/2,d=n/6+1;
40         f(); f2(a);
41         f(); f2(b);
42         f();
43         if(T) printf("\n");
44     }
45     return QAQ;
46 }

hdu 1256

hdu 1257

转化为求最长上升子序列。默写板子...

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 const int maxn=1e6+10;
 6 
 7 int n,a[maxn],dp[maxn];
 8 
 9 int main(){
10     while(scanf("%d",&n)!=EOF){
11         int ans=-1;
12         for(int i=1;i<=n;i++){
13             scanf("%d",&a[i]);
14             dp[i]=1;
15         }
16         for(int i=1;i<=n;i++){
17             for(int j=1;j<i;j++){
18                 if(a[i]>a[j]){
19                     dp[i]=max(dp[j]+1,dp[i]);    
20                 }
21             }
22         }
23         for(int i=1;i<=n;i++){
24             ans=max(ans,dp[i]);
25         }
26         printf("%d\n",ans);
27     }
28     return QAQ;
29 }

hdu 1257

hdu 1258

又是快乐dfs，题目已经排好序了~

注意每次搜的时候去重，比如搜到50的时候，下一个和再下一个都是25，只能搜1次25。也就是同一层不能搜重复数，不然最后会有相同的疯狂刷屏.....

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int t,n,cnt,sum,a[1000];
 7 vector<int>v;
 8 
 9 void dfs(int x){
10     if(sum>t) return;
11     if(sum==t){
12         cnt++;
13         for(int i=0;i<v.size();i++){
14             printf("%d",v[i]);
15             if(i!=v.size()-1) printf("+");
16             else printf("\n");
17         }
18     }
19     
20     int pre=-1;
21     for(int i=x+1;i<=n;i++){
22         if(a[i]==pre) continue;
23         pre=a[i];
24         sum+=a[i];
25         v.push_back(a[i]);
26         dfs(i);
27         sum-=a[i];
28         v.pop_back();
29     }
30 }
31 
32 int main(){
33     while(1){
34         scanf("%d%d",&t,&n);
35         if(t==0 && n==0) break;
36         cnt=0;
37         for(int i=1;i<=n;i++){
38             scanf("%d",&a[i]);
39         }
40         printf("Sums of %d:\n",t);
41         dfs(0);
42         if(!cnt) printf("NONE\n");
43     }
44     return QAQ;
45 }

hdu 1258

hdu 1259

按题意模拟。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int n,m,x,y;
 7 
 8 int main(){
 9     scanf("%d",&n);
10     while(n--){
11         string s="$ZJUTACM";
12         scanf("%d",&m);
13         for(int i=1;i<=m;i++){
14             scanf("%d%d",&x,&y);
15             swap(s[x],s[y]);
16         }
17         for(int i=1;i<=7;i++){
18             if(s[i]=='J'){
19                 printf("%d\n",i);
20                 break;
21             }
22         }
23     }
24     return QAQ;
25 }

hdu 1259

hdu 1260

dp[i]表示让i个人买完票的最短时间。第i个人买票要么单独买，要么和前面的人合买。

状态转移方程dp[i]=min(dp[i-1]+s[i],dp[i-2]+d[i-1])。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int N,k,s[2019],d[2019],dp[2019];
 7 
 8 int main(){
 9     scanf("%d",&N);
10     while(N--){
11         bool f=0;
12         scanf("%d",&k);
13         memset(dp,0,sizeof(dp));
14         for(int i=1;i<=k;i++){
15             scanf("%d",&s[i]);
16         }
17         for(int i=1;i<=k-1;i++){
18             scanf("%d",&d[i]);
19         }
20         
21         dp[1]=s[1];
22         for(int i=2;i<=k;i++){
23             dp[i]=min(dp[i-1]+s[i],dp[i-2]+d[i-1]);
24         }
25         int a=dp[k]/3600+8,b=dp[k]/60%60,c=dp[k]%60;
26         if(a>=12){
27             a-=12;
28             f=1;
29         }
30         printf("%02d:%02d:%02d ",a,b,c);
31         if(f) printf("pm\n");
32         else printf("am\n");
33     }
34     return QAQ;
35 }

hdu 1260

hdu 1261

组合数公式 sigma(a)! / sigma(a!) ，爆ll了，用大数。

  1 #include<bits/stdc++.h>
  2 #define QAQ 0
  3 
  4 using namespace std;
  5 
  6 const int base = 1000000000;
  7 const int base_digits = 9;
  8 
  9 struct Bigint {
 10     vector<int> z;
 11     int sign;
 12 
 13     Bigint() :
 14         sign(1) {
 15     }
 16 
 17     Bigint(long long v) {
 18         *this = v;
 19     }
 20 
 21     Bigint(const string &s) {
 22         read(s);
 23     }
 24 
 25     void operator=(const Bigint &v) {
 26         sign = v.sign;
 27         z = v.z;
 28     }
 29 
 30     void operator=(long long v) {
 31         sign = 1;
 32         if (v < 0)
 33             sign = -1, v = -v;
 34         z.clear();
 35         for (; v > 0; v = v / base)
 36             z.push_back(v % base);
 37     }
 38 
 39     Bigint operator+(const Bigint &v) const {
 40         if (sign == v.sign) {
 41             Bigint res = v;
 42 
 43             for (int i = 0, carry = 0; i < (int) max(z.size(), v.z.size()) || carry; ++i) {
 44                 if (i == (int) res.z.size())
 45                     res.z.push_back(0);
 46                 res.z[i] += carry + (i < (int) z.size() ? z[i] : 0);
 47                 carry = res.z[i] >= base;
 48                 if (carry)
 49                     res.z[i] -= base;
 50             }
 51             return res;
 52         }
 53         return *this - (-v);
 54     }
 55 
 56     Bigint operator-(const Bigint &v) const {
 57         if (sign == v.sign) {
 58             if (abs() >= v.abs()) {
 59                 Bigint res = *this;
 60                 for (int i = 0, carry = 0; i < (int) v.z.size() || carry; ++i) {
 61                     res.z[i] -= carry + (i < (int) v.z.size() ? v.z[i] : 0);
 62                     carry = res.z[i] < 0;
 63                     if (carry)
 64                         res.z[i] += base;
 65                 }
 66                 res.trim();
 67                 return res;
 68             }
 69             return -(v - *this);
 70         }
 71         return *this + (-v);
 72     }
 73 
 74     void operator*=(int v) {
 75         if (v < 0)
 76             sign = -sign, v = -v;
 77         for (int i = 0, carry = 0; i < (int) z.size() || carry; ++i) {
 78             if (i == (int) z.size())
 79                 z.push_back(0);
 80             long long cur = z[i] * (long long) v + carry;
 81             carry = (int) (cur / base);
 82             z[i] = (int) (cur % base);
 83             //asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));
 84         }
 85         trim();
 86     }
 87 
 88     Bigint operator*(int v) const {
 89         Bigint res = *this;
 90         res *= v;
 91         return res;
 92     }
 93 
 94     friend pair<Bigint, Bigint> divmod(const Bigint &a1, const Bigint &b1) {
 95         int norm = base / (b1.z.back() + 1);
 96         Bigint a = a1.abs() * norm;
 97         Bigint b = b1.abs() * norm;
 98         Bigint q, r;
 99         q.z.resize(a.z.size());
100 
101         for (int i = a.z.size() - 1; i >= 0; i--) {
102             r *= base;
103             r += a.z[i];
104             int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
105             int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
106             int d = ((long long) s1 * base + s2) / b.z.back();
107             r -= b * d;
108             while (r < 0)
109                 r += b, --d;
110             q.z[i] = d;
111         }
112 
113         q.sign = a1.sign * b1.sign;
114         r.sign = a1.sign;
115         q.trim();
116         r.trim();
117         return make_pair(q, r / norm);
118     }
119 
120     friend Bigint sqrt(const Bigint &a1) {
121         Bigint a = a1;
122         while (a.z.empty() || a.z.size() % 2 == 1)
123             a.z.push_back(0);
124 
125         int n = a.z.size();
126 
127         int firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]);
128         int norm = base / (firstDigit + 1);
129         a *= norm;
130         a *= norm;
131         while (a.z.empty() || a.z.size() % 2 == 1)
132             a.z.push_back(0);
133 
134         Bigint r = (long long) a.z[n - 1] * base + a.z[n - 2];
135         firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]);
136         int q = firstDigit;
137         Bigint res;
138 
139         for(int j = n / 2 - 1; j >= 0; j--) {
140             for(; ; --q) {
141                 Bigint r1 = (r - (res * 2 * base + q) * q) * base * base + (j > 0 ? (long long) a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0);
142                 if (r1 >= 0) {
143                     r = r1;
144                     break;
145                 }
146             }
147             res *= base;
148             res += q;
149 
150             if (j > 0) {
151                 int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
152                 int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
153                 int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
154                 q = ((long long) d1 * base * base + (long long) d2 * base + d3) / (firstDigit * 2);
155             }
156         }
157 
158         res.trim();
159         return res / norm;
160     }
161 
162     Bigint operator/(const Bigint &v) const {
163         return divmod(*this, v).first;
164     }
165 
166     Bigint operator%(const Bigint &v) const {
167         return divmod(*this, v).second;
168     }
169 
170     void operator/=(int v) {
171         if (v < 0)
172             sign = -sign, v = -v;
173         for (int i = (int) z.size() - 1, rem = 0; i >= 0; --i) {
174             long long cur = z[i] + rem * (long long) base;
175             z[i] = (int) (cur / v);
176             rem = (int) (cur % v);
177         }
178         trim();
179     }
180 
181     Bigint operator/(int v) const {
182         Bigint res = *this;
183         res /= v;
184         return res;
185     }
186 
187     int operator%(int v) const {
188         if (v < 0)
189             v = -v;
190         int m = 0;
191         for (int i = z.size() - 1; i >= 0; --i)
192             m = (z[i] + m * (long long) base) % v;
193         return m * sign;
194     }
195 
196     void operator+=(const Bigint &v) {
197         *this = *this + v;
198     }
199     void operator-=(const Bigint &v) {
200         *this = *this - v;
201     }
202     void operator*=(const Bigint &v) {
203         *this = *this * v;
204     }
205     void operator/=(const Bigint &v) {
206         *this = *this / v;
207     }
208 
209     bool operator<(const Bigint &v) const {
210         if (sign != v.sign)
211             return sign < v.sign;
212         if (z.size() != v.z.size())
213             return z.size() * sign < v.z.size() * v.sign;
214         for (int i = z.size() - 1; i >= 0; i--)
215             if (z[i] != v.z[i])
216                 return z[i] * sign < v.z[i] * sign;
217         return false;
218     }
219 
220     bool operator>(const Bigint &v) const {
221         return v < *this;
222     }
223     bool operator<=(const Bigint &v) const {
224         return !(v < *this);
225     }
226     bool operator>=(const Bigint &v) const {
227         return !(*this < v);
228     }
229     bool operator==(const Bigint &v) const {
230         return !(*this < v) && !(v < *this);
231     }
232     bool operator!=(const Bigint &v) const {
233         return *this < v || v < *this;
234     }
235 
236     void trim() {
237         while (!z.empty() && z.back() == 0)
238             z.pop_back();
239         if (z.empty())
240             sign = 1;
241     }
242 
243     bool isZero() const {
244         return z.empty() || (z.size() == 1 && !z[0]);
245     }
246 
247     Bigint operator-() const {
248         Bigint res = *this;
249         res.sign = -sign;
250         return res;
251     }
252 
253     Bigint abs() const {
254         Bigint res = *this;
255         res.sign *= res.sign;
256         return res;
257     }
258 
259     long long longValue() const {
260         long long res = 0;
261         for (int i = z.size() - 1; i >= 0; i--)
262             res = res * base + z[i];
263         return res * sign;
264     }
265 
266     friend Bigint gcd(const Bigint &a, const Bigint &b) {
267         return b.isZero() ? a : gcd(b, a % b);
268     }
269     friend Bigint lcm(const Bigint &a, const Bigint &b) {
270         return a / gcd(a, b) * b;
271     }
272 
273     void read(const string &s) {
274         sign = 1;
275         z.clear();
276         int pos = 0;
277         while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) {
278             if (s[pos] == '-')
279                 sign = -sign;
280             ++pos;
281         }
282         for (int i = s.size() - 1; i >= pos; i -= base_digits) {
283             int x = 0;
284             for (int j = max(pos, i - base_digits + 1); j <= i; j++)
285                 x = x * 10 + s[j] - '0';
286             z.push_back(x);
287         }
288         trim();
289     }
290 
291     friend istream& operator>>(istream &stream, Bigint &v) {
292         string s;
293         stream >> s;
294         v.read(s);
295         return stream;
296     }
297 
298     friend ostream& operator<<(ostream &stream, const Bigint &v) {
299         if (v.sign == -1)
300             stream << '-';
301         stream << (v.z.empty() ? 0 : v.z.back());
302         for (int i = (int) v.z.size() - 2; i >= 0; --i)
303             stream << setw(base_digits) << setfill('0') << v.z[i];
304         return stream;
305     }
306 
307     static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
308         vector<long long> p(max(old_digits, new_digits) + 1);
309         p[0] = 1;
310         for (int i = 1; i < (int) p.size(); i++)
311             p[i] = p[i - 1] * 10;
312         vector<int> res;
313         long long cur = 0;
314         int cur_digits = 0;
315         for (int i = 0; i < (int) a.size(); i++) {
316             cur += a[i] * p[cur_digits];
317             cur_digits += old_digits;
318             while (cur_digits >= new_digits) {
319                 res.push_back(int(cur % p[new_digits]));
320                 cur /= p[new_digits];
321                 cur_digits -= new_digits;
322             }
323         }
324         res.push_back((int) cur);
325         while (!res.empty() && res.back() == 0)
326             res.pop_back();
327         return res;
328     }
329 
330     typedef vector<long long> vll;
331 
332     static vll karatsubaMultiply(const vll &a, const vll &b) {
333         int n = a.size();
334         vll res(n + n);
335         if (n <= 32) {
336             for (int i = 0; i < n; i++)
337                 for (int j = 0; j < n; j++)
338                     res[i + j] += a[i] * b[j];
339             return res;
340         }
341 
342         int k = n >> 1;
343         vll a1(a.begin(), a.begin() + k);
344         vll a2(a.begin() + k, a.end());
345         vll b1(b.begin(), b.begin() + k);
346         vll b2(b.begin() + k, b.end());
347 
348         vll a1b1 = karatsubaMultiply(a1, b1);
349         vll a2b2 = karatsubaMultiply(a2, b2);
350 
351         for (int i = 0; i < k; i++)
352             a2[i] += a1[i];
353         for (int i = 0; i < k; i++)
354             b2[i] += b1[i];
355 
356         vll r = karatsubaMultiply(a2, b2);
357         for (int i = 0; i < (int) a1b1.size(); i++)
358             r[i] -= a1b1[i];
359         for (int i = 0; i < (int) a2b2.size(); i++)
360             r[i] -= a2b2[i];
361 
362         for (int i = 0; i < (int) r.size(); i++)
363             res[i + k] += r[i];
364         for (int i = 0; i < (int) a1b1.size(); i++)
365             res[i] += a1b1[i];
366         for (int i = 0; i < (int) a2b2.size(); i++)
367             res[i + n] += a2b2[i];
368         return res;
369     }
370 
371     Bigint operator*(const Bigint &v) const {
372         vector<int> a6 = convert_base(this->z, base_digits, 6);
373         vector<int> b6 = convert_base(v.z, base_digits, 6);
374         vll a(a6.begin(), a6.end());
375         vll b(b6.begin(), b6.end());
376         while (a.size() < b.size())
377             a.push_back(0);
378         while (b.size() < a.size())
379             b.push_back(0);
380         while (a.size() & (a.size() - 1))
381             a.push_back(0), b.push_back(0);
382         vll c = karatsubaMultiply(a, b);
383         Bigint res;
384         res.sign = sign * v.sign;
385         for (int i = 0, carry = 0; i < (int) c.size(); i++) {
386             long long cur = c[i] + carry;
387             res.z.push_back((int) (cur % 1000000));
388             carry = (int) (cur / 1000000);
389         }
390         res.z = convert_base(res.z, 6, base_digits);
391         res.trim();
392         return res;
393     }
394 };
395 
396 int n;
397 
398 int main(){
399     while(scanf("%d",&n)){
400         if(n==0) break;
401         Bigint ans=1;
402         int a[30],cnt=0;
403         for(int i=1;i<=n;i++){
404             cin>>a[i];
405             cnt+=a[i];
406         }
407         for(int i=1;i<=cnt;i++){
408             ans*=i;
409         }
410         for(int i=1;i<=n;i++){
411             for(int j=1;j<=a[i];j++){
412                 ans/=j;
413             }
414         }
415         cout<<ans<<endl;
416     }
417     return QAQ;
418 }

hdu 1261

hdu 1262

线性筛预处理一下素数，从m/2开始搜，第一个搜到的一定是最近的。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 const int maxn=100000;
 6 int prime[maxn],check[maxn];
 7 int m,n;
 8 
 9 void xxs(){
10     int tot=0;
11     memset(check,0,sizeof(check));
12     check[1]=1; 
13     for(int i=2;i<maxn;i++){
14         if(!check[i]) prime[tot++]=i;
15         for(int j=0;j<tot;j++){
16             if(i*prime[j]>maxn) break;
17             check[i*prime[j]]=1;
18             if(i%prime[j]==0) break;
19         }
20     }
21 } 
22 
23 int main(){
24     xxs();
25     while(scanf("%d",&m)!=EOF){
26         n=m/2;
27         while(1){
28             if(!check[n] && !check[m-n]){
29                 printf("%d %d\n",n,m-n);
30                 break;
31             }
32             n--;
33         }
34     }
35     return QAQ;
36 }

hdu 1262

hdu 1263

二维map计数。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int N,M,a;
 7 map<string,map<string,int> >m;
 8 string s1,s2;
 9 
10 int main(){
11     cin>>N;
12     while(N--){
13         m.clear();
14         cin>>M;
15         for(int i=1;i<=M;i++){
16             cin>>s1>>s2>>a;
17             m[s2][s1]+=a;        
18         }
19         for(auto x:m){
20             cout<<x.first<<'\n';
21             for(auto y:x.second){
22                 cout<<"   |----"<<y.first<<'('<<y.second<<')'<<'\n';
23             }
24         }
25         if(N) cout<<'\n'; 
26     }
27     return QAQ;
28 }

hdu 1263

hdu 1264

注意数据范围0到100...直接暴力扫。确定矩形位置的两个点有两种情况，一开始没swap掉坑里了。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int a1,b1,a2,b2;
 7 bool m[102][102];
 8 
 9 int main(){
10     while(1){
11         memset(m,0,sizeof(m));
12         int cnt=0;
13         while(1){
14             scanf("%d%d%d%d",&a1,&b1,&a2,&b2);
15             if(a1==-1 || a1==-2) break;
16             if(a1>a2) swap(a1,a2);
17             if(b1>b2) swap(b1,b2);
18             
19             for(int i=b1;i<b2;i++){
20                 for(int j=a1;j<a2;j++){
21                     if(!m[i][j]){
22                         cnt++;
23                         m[i][j]=1;
24                     }
25                 }
26             }
27         }
28         printf("%d\n",cnt);
29         if(a1==-2) break;
30     }
31     return QAQ;
32 }

hdu 1264

hdu 1265

十六进制表示浮点数...挺奇特的...完全没想到这个思路。直接把计算机存放的二进制化成十六进制。。。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int N;
 7 float f;
 8 
 9 int main(){
10     scanf("%d",&N);
11     while(N--){
12         scanf("%f",&f);
13         unsigned char *p=(unsigned char *)&f;
14         printf("%02X%02X%02X%02X\n",p[3],p[2],p[1],p[0]);
15     }
16     return QAQ;
17 }

hdu 1265

hdu 1491

模拟题。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 
 6 int T,a,b;
 7 int m[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 8 
 9 int main(){
10     scanf("%d",&T);
11     while(T--){
12         scanf("%d%d",&a,&b);
13         if(a>10 || (a==10&&b>21)){
14             printf("What a pity, it has passed!\n");
15         }
16         else if(a==10 && b==21){
17             printf("It's today!!\n");
18         }
19         else{
20             if(a==10){
21                 printf("%d\n",21-b);
22             }
23             else{
24                 int ans=21+m[a]-b;
25                 for(int i=a+1;i<=9;i++){
26                     ans+=m[i];
27                 }
28                 printf("%d\n",ans);
29             }
30         }
31     }
32     return QAQ;
33 }

hdu 1491

hdu 1492

因子只有2，3，5，7。统计它们的幂次，用求因数个数的公式就好。不开ll会wa。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 
 4 using namespace std;
 5 typedef long long ll;
 6 
 7 ll n;
 8 ll f[4]={2,3,5,7},cnt[4];
 9 
10 int main(){
11     while(1){
12         scanf("%lld",&n);
13         if(n==0) break;
14         ll ans=1;
15         for(int i=0;i<4;i++){
16             ll t=n;
17             cnt[i]=0;
18             while(t%f[i]==0){
19                 cnt[i]++;
20                 t/=f[i];
21             }
22             ans*=(cnt[i]+1);
23         }
24         printf("%lld\n",ans);
25     }
26     return QAQ;
27 }

hdu 1492

hdu 1493

概率dp。本来想开二维，其实可以滚动，每次的概率以上次为基础。每次从后往前，dp[i]=dp[i]+dp[i-k]*rp[k]，k为掷出的点数。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 #define miaojie1
 4 #ifdef miaojie
 5   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
 6 void err() {std::cout << std::endl;}
 7 template<typename T, typename...Args>
 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
 9 #else
10   #define dbg(...)
11 #endif
12 
13 using namespace std;
14 
15 int T;
16 double rp[10],dp[100];
17 int a[11]={0,5,12,22,29,33,38,42,46,50,55};
18 
19 int main(){
20     scanf("%d",&T);
21     
22     while(T--){
23         for(int i=1;i<=6;i++){
24             scanf("%lf",&rp[i]);
25         }
26         memset(dp,0,sizeof(dp));
27         dp[0]=1;
28         
29         for(int i=1;i<=10;i++){
30             dbg(dp[12]);
31             for(int j=60;j>0;j--){
32                 dp[j]=0;
33                 for(int k=1;k<=6;k++){
34                     if(j<k) break;
35                     dp[j]=dp[j]+rp[k]*dp[j-k];
36                 }
37             }
38         }
39         for(int i=1;i<=10;i++){
40             printf("%d: %.1lf%%\n",a[i],100.0*dp[a[i]]);
41         }
42         if(T) printf("\n");
43     }
44     return QAQ;
45 }

hdu 1493

hdu 1494

还是dp。把n圈展开成线，能量看成0-14的整数，dp[i][j]表示到第i段时有j格能量所用的最短时间，顺序dp。注意j=0一定是刚用过加速卡，j=10可能是普通行驶或爆卡，j>10一定是普通行驶...分类挺麻烦的，状态方程分四个。感觉还可以简化下。

 1 #include<bits/stdc++.h>
 2 #define QAQ 0
 3 #define miaojie1
 4 #ifdef miaojie
 5   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
 6 void err() {std::cout << std::endl;}
 7 template<typename T, typename...Args>
 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
 9 #else
10   #define dbg(...)
11 #endif
12 
13 using namespace std;
14 typedef long long ll;
15 
16 int l,n;
17 ll a[10020],b[10020],dp[10020][15];
18 
19 int main(){
20     while(scanf("%d",&l)!=EOF){
21         scanf("%d",&n);
22         int sum=n*l;
23         for(int i=1;i<=l;i++){
24             scanf("%d",&a[i]);
25             for(int j=1;j<=n-1;j++){
26                 a[i+j*l]=a[i];
27             }
28         }
29         for(int i=1;i<=l;i++){
30             scanf("%d",&b[i]);
31             for(int j=1;j<=n-1;j++){
32                 b[i+j*l]=b[i];
33             }
34         }
35         for(int i=0;i<=sum;i++){
36             for(int j=0;j<15;j++){
37                 dp[i][j]=1ll<<30;
38             }
39         }
40         dp[0][0]=0;
41         
42         for(int i=1;i<=sum;i++){
43             for(int j=0;j<15;j++){
44                 if(j==0) {
45                     dp[i][j]=dp[i-1][5]+b[i];    
46                 }
47                 else if(j==10){
48                     dp[i][j]=min( dp[i-1][14]+a[i] , dp[i-1][9]+a[i] );
49                 }
50                 else if(j>10){
51                     dp[i][j]=dp[i-1][j-1]+a[i];
52                 }
53                 else dp[i][j]=min( dp[i-1][j-1]+a[i] , dp[i-1][j+5]+b[i] );
54             }
55         }
56         ll ans=1ll<<62;
57         for(int i=0;i<15;i++){
58             ans=min(ans,dp[sum][i]);
59             dbg(dp[sum][i]);
60         }
61         printf("%lld\n",ans);
62     }
63     return QAQ;
64 }

hdu 1494

hdu 1495

bfs，模拟每步的六种操作即可。

  1 #include<bits/stdc++.h>
  2 #define miaojie1
  3 #ifdef miaojie
  4   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
  5 void err() {std::cout << std::endl;}
  6 template<typename T, typename...Args>
  7 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
  8 #else
  9   #define dbg(...)
 10 #endif
 11 
 12 using namespace std;
 13 
 14 int S,s,n,m;
 15 bool vis[105][105][105];
 16 
 17 struct node{
 18     int a,b,c,r;
 19     node(){}
 20     node(int _a,int _b,int _c,int _r) : a(_a),b(_b),c(_c),r(_r){}
 21 };
 22 
 23 void bfs(){
 24     memset(vis,0,sizeof(vis));
 25     if(S%2){
 26         printf("NO\n");
 27         return;
 28     }
 29     else s=S/2;
 30     queue<node>q;
 31     node st(S,0,0,0);
 32     q.push(st);
 33     vis[s][0][0]=1;
 34     
 35     while(!q.empty()){
 36         node t=q.front();
 37         dbg(t.a,t.b,t.c,t.r);
 38         if( (t.a==s&&t.b==s) || (t.a==s&&t.c==s) || (t.b==s&&t.c==s)){
 39             printf("%d\n",t.r);
 40             return;
 41         }
 42         q.pop();
 43         if(t.a>0){
 44             node p=t;
 45             if(p.a>=n-p.b){
 46                 p.a=p.a-(n-p.b);
 47                 p.b=n;
 48                 p.r++;
 49             }
 50             else{
 51                 p.b+=p.a;
 52                 p.a=0;
 53                 p.r++;
 54             }
 55             if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1;
 56             
 57             p=t;
 58             if(p.a>=m-p.c){
 59                 p.a=p.a-(m-p.c);
 60                 p.c=m;
 61                 p.r++;
 62             }
 63             else{
 64                 p.c+=p.a;
 65                 p.a=0;
 66                 p.r++;
 67             }
 68             if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1;
 69             
 70             p=t;
 71             if(p.b>=S-p.a){
 72                 p.b=p.b-(S-p.a);
 73                 p.a=s;
 74                 p.r++;
 75             }
 76             else{
 77                 p.a+=p.b;
 78                 p.b=0;
 79                 p.r++;
 80             }
 81             if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1;
 82             
 83             p=t;
 84             if(p.b>=m-p.c){
 85                 p.b=p.b-(m-p.c);
 86                 p.c=m;
 87                 p.r++;
 88             }
 89             else{
 90                 p.c+=p.b;
 91                 p.b=0;
 92                 p.r++;
 93             }
 94             if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1;
 95             
 96             p=t;
 97             if(p.c>=S-p.a){
 98                 p.c=p.c-(S-p.a);
 99                 p.a=s;
100                 p.r++;
101             }
102             else{
103                 p.a+=p.c;
104                 p.c=0;
105                 p.r++;
106             }
107             if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1;
108             
109             p=t;
110             if(p.c>=n-p.b){
111                 p.c=p.c-(n-p.b);
112                 p.b=n;
113                 p.r++;
114             }
115             else{
116                 p.b+=p.c;
117                 p.c=0;
118                 p.r++;
119             }
120             if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1;
121         }
122     }
123     printf("NO\n");
124 }
125 
126 int main(){
127     while(1){
128         scanf("%d%d%d",&S,&n,&m);
129         if(S==0 && n==0 && m==0) break;
130         bfs();
131     }
132     return 0;
133 }

hdu 1495

hdu 1496

四个for显然超了，用折半法降时间复杂度，结果还是T，加了特判才过。由于有平方，只讨论正号省时间，最后乘16就行。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 
 6 int a,b,c,d;
 7 map<int,int>mp;
 8 
 9 int main(){
10     while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF){
11         if((a>0 && b>0 && c>0 && d>0) || (a<0 && b<0 && c<0 && d<0)){
12             printf("0\n");
13             continue;
14         } 
15         ll ans=0;
16         mp.clear();
17         for(int i=1;i<=100;i++){
18             for(int j=1;j<=100;j++){
19                 mp[a*i*i+b*j*j]++;
20             }
21         }
22         for(int i=1;i<=100;i++){
23             for(int j=1;j<=100;j++){
24                 ans+=mp[-(c*i*i+d*j*j)];
25             }
26         }
27         ans*=16;
28         printf("%lld\n",ans);
29     }
30     return 0;
31 }

hdu 1496

hdu 1497

怎么又是大模拟，8想做！不看它了！

hdu 1498

行对列匹配跑一遍二分图的最小覆盖,balabala（明天补上）

第二天：好像也补不上啥啊。。多一个匹配就多一个既不同行也不同列的气球，超过k个就不行了。

 1 #include<bits/stdc++.h>
 2 #define miaojie1
 3 #ifdef miaojie
 4   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
 5 void err() {std::cout << std::endl;}
 6 template<typename T, typename...Args>
 7 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
 8 #else
 9   #define dbg(...)
10 #endif
11 using namespace std;
12 
13 int n,k,a[105][105],edge[105][105],cx[105],cy[105],ans[105],pp;
14 bool vis[105];
15 set<int>se;
16 
17 int line(int u){
18     for(int v=1;v<=n;v++){
19         if(edge[u][v] && !vis[v]){
20             vis[v]=1;
21             if(cy[v]==-1 || line(cy[v])){
22                 cx[u]=v;
23                 cy[v]=u;
24                 return 1;
25             }
26         }
27     }
28     return 0;
29 }
30 
31 int mincover(){
32     int ret=0;
33     memset(cx,-1,sizeof(cx));
34     memset(cy,-1,sizeof(cy));
35     for(int i=1;i<=n;i++){
36         if(cx[i]==-1){
37             memset(vis,0,sizeof(vis));
38             ret+=line(i);
39         }
40     }
41     return ret;
42 }
43 
44 int main(){
45     while(1){
46         scanf("%d%d",&n,&k);
47         se.clear(); pp=0;
48         if(n==0 && k==0) break;
49         for(int i=1;i<=n;i++){
50             for(int j=1;j<=n;j++){
51                 scanf("%d",&a[i][j]);
52                 se.insert(a[i][j]);
53             }
54         }
55         for(auto x:se){
56             dbg(x); 
57             memset(edge,0,sizeof(edge));
58             for(int i=1;i<=n;i++){
59                 for(int j=1;j<=n;j++){
60                     if(a[i][j]==x){
61                         edge[i][j]=1;
62                     }
63                 }
64             }
65             if(mincover()>k) ans[pp++]=x;
66         }
67         dbg(pp);
68         if(pp==0){
69             printf("-1\n");
70         }
71         else{
72             for(int i=0;i<pp;i++){
73                 printf("%d",ans[i]);
74                 if(i!=pp-1) printf(" ");
75                 else printf("\n");
76             }
77         }
78     }
79     return 0;
80 }

hdu 1498

hdu 1728

一开始传统bfs，出队一个就搜四周，记录朝向来表示是否转弯，mle了。

其实这题应该换个想法，答案只和拐弯数有关，所以搜的时候可以不再前进一个点，改为一直走到某方向的尽头，路上的点还是判vis按顺序入队。这样每个点搜出一个十字形状，并且保证了第一次搜到时拐弯数最小。

定义了y1，没敢用万能头文件。

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<string.h>
 7 using namespace std;
 8 
 9 int T,m,n;
10 char a[105][105];
11 bool vis[105][105];
12 int k,x1,x2,y1,y2;
13 int tx[5]={0,-1,0,1,0},ty[5]={0,0,1,0,-1};
14 
15 struct node{
16     int x,y,r;
17     node(){}
18     node(int _x,int _y,int _r) : x(_x),y(_y),r(_r){}
19 };
20 
21 void bfs(){
22     queue<node>q;
23     vis[x1][y1]=1;
24 //    node st1(x1,y1,-1,1),st2(x1,y1,-1,2),st3(x1,y1,-1,3),st4(x1,y1,-1,4);
25 //    q.push(st1); q.push(st2); q.push(st3); q.push(st4);
26     node st(x1,y1,-1);
27     q.push(st);
28     while(!q.empty()){
29         node t=q.front();
30     //    cout<<"!!!"<<t.x<<" "<<t.y<<" "<<t.r<<" "<<endl;
31         q.pop();
32         if(t.r>k) continue;
33         if(t.x==x2 && t.y==y2){
34     //        cout<<"~~~"<<t.r<<" "<<t.s<<endl;
35             printf("yes\n");
36             return;
37         }
38         for(int i=1;i<=4;i++){
39             node p=t;
40             p.x+=tx[i];
41             p.y+=ty[i];
42             p.r++;
43             while(!(p.x<1||p.x>m||p.y<1||p.y>n||p.r>k||a[p.x][p.y]=='*')){    
44                 if(!vis[p.x][p.y]){
45                     vis[p.x][p.y]=1;
46                     q.push(p);
47                 }
48                 p.x+=tx[i];
49                 p.y+=ty[i];
50             }
51         }
52     }
53     printf("no\n");
54 }
55 
56 int main(){
57     scanf("%d",&T);
58     while(T--){
59         scanf("%d%d",&m,&n);
60         memset(vis,0,sizeof(vis));
61         for(int i=1;i<=m;i++){
62             for(int j=1;j<=n;j++){
63                 cin>>a[i][j];
64             }
65         }
66         scanf("%d%d%d%d%d",&k,&y1,&x1,&y2,&x2);
67         bfs();
68     } 
69     return 0;
70 }

hdu 1728

hdu 1729

sg函数太生疏了.....

s为容量，找到临界点t+t*t<s且(t+1)+(t+1)^2>=s，可知当盒子里大于t个时必胜，返回容量减现有个数（类比取石子），等于t时必败。小于t时无法判断，但s和t都是必败点，返回（t，c）等价。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int n,cnt,s,c;
 6 
 7 int SG(int s,int c){
 8     int t=sqrt(s);
 9     while(t+t*t>=s){
10         t--;
11     }
12     if(c>t) return s-c;
13     else return SG(t,c);
14 }
15 
16 int main(){
17     while(1){
18         scanf("%d",&n);
19         if(n==0) break;
20         int ans=0;
21         for(int i=1;i<=n;i++){
22             scanf("%d%d",&s,&c);
23             ans^=SG(s,c);
24         }
25         printf("Case %d:\n",++cnt);
26         if(ans) printf("Yes\n");
27         else printf("No\n");
28     }
29     return 0;
30 }

hdu 1729

hdu 1730

这题友好多了，每行两个棋子的距离-1其实就类比每堆石子数，nim一下就好。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int n,m,a,b;
 6 
 7 int main(){
 8     while(scanf("%d%d",&n,&m)!=EOF){
 9         int ans=0;
10         for(int i=1;i<=n;i++){
11             scanf("%d%d",&a,&b);
12             ans=ans^(abs(a-b)-1);
13         }
14         if(!ans){
15             printf("BAD LUCK!\n");
16         }
17         else printf("I WIN!\n");
18     }
19     return 0;
20 }

hdu 1730

hdu 1732

bfs，记录人和三个箱子的位置。走一步——是否有障碍或箱子——若有箱子，箱子后面是否有障碍或其他箱子。

因为把=写成==爆内存了，找了一个小时，强行拉底ac率，很烦。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 struct node{
 6     int x[4],y[4],r;
 7 }st;
 8 
 9 int n,m;
10 int tx[4]={-1,0,1,0},ty[4]={0,1,0,-1},endx[4],endy[4];
11 char a[8][8];
12 bool vis[8][8][8][8][8][8][8][8];
13 queue<node>q;
14 void bfs(){
15     memset(vis,0,sizeof(vis));
16     while(!q.empty()) q.pop();
17     q.push(st);
18     while(!q.empty()){
19         node t=q.front();
20         q.pop();
21         bool check=1;
22         for(int i=1;i<=3;i++){
23             bool ck=0;
24             for(int j=1;j<=3;j++){
25                 if(t.x[i]==endx[j] && t.y[i]==endy[j]) ck=1;
26             }
27             if(!ck){
28                 check=0;
29                 break;
30             }
31         }
32         if(check){
33             printf("%d\n",t.r);
34             return;
35         }
36         if(vis[t.x[0]][t.y[0]][t.x[1]][t.y[1]][t.x[2]][t.y[2]][t.x[3]][t.y[3]]==1) continue;
37         vis[t.x[0]][t.y[0]][t.x[1]][t.y[1]][t.x[2]][t.y[2]][t.x[3]][t.y[3]]==1;
38         for(int i=0;i<=3;i++){
39             node p=t;
40             int f=0;
41             p.x[0]+=tx[i];
42             p.y[0]+=ty[i];
43             p.r++;
44             if(a[p.x[0]][p.y[0]]=='#'||p.x[0]>=n||p.x[0]<0||p.y[0]>=m||p.y[0]<0) continue;
45             for(int j=1;j<=3;j++){
46                 if(p.x[0]==p.x[j] && p.y[0]==p.y[j]){
47                     f=j;
48                     break;
49                 }
50             }
51             if(f){
52                 p.x[f]+=tx[i];
53                 p.y[f]+=ty[i];
54                 if(a[p.x[f]][p.y[f]]=='#'||p.x[f]>=n||p.x[f]<0||p.y[f]>=m||p.y[f]<0) continue;
55                 bool ff=0;
56                 for(int j=1;j<=3;j++){
57                     if(j==f) continue;
58                     if(p.x[f]==p.x[j] && p.y[f]==p.y[j]){
59                         ff=1;
60                         break;
61                     }
62                 }
63                 if(!ff) q.push(p);
64             }
65             else{
66                 q.push(p);
67             }
68         }
69     }
70     printf("-1\n");
71 }
72 
73 int main(){
74     ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);
75     while(cin>>n>>m){
76         int p=0,q=0;
77         for(int i=0;i<n;i++){
78             for(int j=0;j<m;j++){
79                 cin>>a[i][j];
80                 if(a[i][j]=='X'){
81                     st.x[0]=i;
82                     st.y[0]=j;
83                 }
84                 else if(a[i][j]=='*'){
85                     st.x[++p]=i;
86                     st.y[p]=j;
87                 }
88                 else if(a[i][j]=='@'){
89                     endx[++q]=i;
90                     endy[q]=j;
91                 }
92             }
93         }
94         st.r=0;
95         bfs();
96     }
97     return 0;
98 }

hdu 1732

hdu 5702

排序。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 struct node{
 6     char s[20];
 7     int v;
 8     bool operator < (const node &t) const{
 9         return v>t.v;
10     }
11 }a[20];
12 
13 int T,n;
14 
15 int main(){
16     scanf("%d",&T);
17     while(T--){
18         scanf("%d",&n);
19         for(int i=1;i<=n;i++){
20             scanf("%s%d",a[i].s,&a[i].v);
21         }
22         sort(a+1,a+1+n);
23         for(int i=1;i<=n;i++){
24             printf("%s",a[i].s);
25             if(i==n) printf("\n");
26             else printf(" ");
27         }
28     }
29     return 0;
30 }

hdu 5702

hdu 5703

列出前几项就能发现规律。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int T,n;
 6 
 7 int main(){
 8     scanf("%d",&T);
 9     while(T--){
10         scanf("%d",&n);
11         printf("1");
12         for(int i=1;i<=n-1;i++){
13             printf("0");
14         }
15         printf("\n");
16     }
17     return 0;
18 }

hdu 5703

hdu 5704

方程列出来，解方程。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int T,n,sum,a[102];
 6 map<int,int>m;
 7 
 8 int main(){
 9     scanf("%d",&T);
10     while(T--){
11         m.clear(); sum=0;
12         scanf("%d",&n);
13         for(int i=1;i<=n-1;i++){
14             scanf("%d",&a[i]);
15             m[a[i]]++;
16             sum+=a[i];
17         }
18         int ans=2*sum/(3*n-2);
19         double p=1.0/(m[ans]+1.0);
20         printf("%d %.2lf\n",ans,p);
21     }
22     return 0;
23 }

hdu 5704

hdu 5705

计算时针和分针的移动速度，从00：00：00开始枚举判断角度。为了避开精度问题乘了11。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int h,m,s,a,cas,ansh,ansm,anss;
 6 
 7 int main(){
 8     ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
 9     while(~scanf("%d:%d:%d%d",&h,&m,&s,&a)){
10         printf("Case #%d: ",++cas);
11         int t=(h*3600+m*60+s)*11,tim=0,cnt=0;
12         while(1){
13             tim+=120;
14             cnt+=11;
15             int tmp=cnt%3960;
16             if(tmp>1980) tmp=3960-tmp;
17             if(tmp==a*11&&tim>t) break;
18         }
19         tim/=11;
20         tim=tim%(12*60*60);
21         ansh=tim/3600;
22         ansm=(tim-ansh*3600)/60;
23         anss=tim-ansh*3600-ansm*60;
24         printf("%02d:%02d:%02d\n",ansh,ansm,anss);
25     }
26     return 0;
27 }

hdu 5705

hdu 5706

dfs连通块，搜就行了。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int T,n,m,ans1,ans2;
 6 char a[1005][1005];
 7 char s1[10]="girl",s2[10]="cat";
 8 
 9 void dfs1(int i,int j,int k){
10     if(i<1||i>n||j<1||j>m) return;
11     if(a[i][j]==s1[k]){
12         k++;
13         if(k==4){
14             ans1++;
15             return;
16         }
17         dfs1(i-1,j,k);
18         dfs1(i+1,j,k);
19         dfs1(i,j+1,k);
20         dfs1(i,j-1,k); 
21     }    
22 }
23 
24 void dfs2(int i,int j,int k){
25     if(i<1||i>n||j<1||j>m) return;
26     if(a[i][j]==s2[k]){
27         k++;
28         if(k==3){
29             ans2++;
30             return;
31         }
32         dfs2(i-1,j,k);
33         dfs2(i+1,j,k);
34         dfs2(i,j+1,k);
35         dfs2(i,j-1,k); 
36     }    
37 }
38 
39 int main(){
40     scanf("%d",&T);
41     while(T--){
42         ans1=ans2=0;
43         scanf("%d%d",&n,&m);
44         for(int i=1;i<=n;i++){
45             scanf("%s",a[i]+1);
46         }
47         for(int i=1;i<=n;i++){
48             for(int j=1;j<=m;j++){
49                 if(a[i][j]=='g') dfs1(i,j,0);
50                 if(a[i][j]=='c') dfs2(i,j,0);
51             }
52         }
53         printf("%d %d\n",ans1,ans2);
54     }
55     return 0;
56 }

hdu 5706

hdu 5707

由于c中所有字母都要使用，按顺序对于每个字母判断是否能加入即可。dp两维分别表示a和b能匹配的字母数。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 bool dp[2002][2002];
 6 char a[2002],b[2002],c[2002];
 7 
 8 int main(){
 9     while(~scanf("%s%s%s",a,b,c)){
10         int la=strlen(a),lb=strlen(b),lc=strlen(c);
11         if(lc!=la+lb){
12             printf("No\n");
13             continue;
14         }
15         memset(dp,0,sizeof(dp));
16         dp[0][0]=1;
17         for(int i=0;i<=la;i++){
18             for(int j=0;j<=lb;j++){
19                 if(c[i+j]==a[i]){
20                     dp[i+1][j]|=dp[i][j];
21                 }
22                 if(c[i+j]==b[j]){
23                     dp[i][j+1]|=dp[i][j];
24                 }
25             }
26         }
27         if(dp[la][lb]) printf("Yes\n");
28         else printf("No\n");
29     }
30     return 0;
31 }

hdu 5707

hdu5708

sg打表找规律.....看了半天，周期居然是2*（k+1）。k=1需要特判。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 const int N=30;
 5 
 6 int T,q,k,n,m;
 7 int sg[50][50];
 8 
 9 void get_SG(){
10     sg[1][1]=0;
11     for(int i=2;i<=N;i+=2){
12         sg[i][1]=sg[1][i]=1;
13         sg[i+1][1]=sg[1][i+1]=0;
14     }
15     for(int i=2;i<=N;i++){
16         for(int j=2;j<=N;j++){
17             if(!sg[i-1][j] || !sg[i][j-1] || (i-k>0 && j-k>0 && (!sg[i-k][j-k])) )
18                 sg[i][j]=1;
19             else sg[i][j]=0;
20         }
21     }
22 }
23 
24 void p1(){
25     printf("Alice\n");
26 }
27 
28 void p2(){
29     printf("Bob\n");
30 }
31 
32 int main(){
33     scanf("%d",&T);
34     while(T--){
35         scanf("%d%d",&q,&k);
36     /*    memset(sg,0,sizeof(sg));
37         get_SG();
38         printf("!!!\n");
39         for(int i=1;i<=N;i++){
40             for(int j=1;j<=N;j++){
41                 printf("%d ",sg[i][j]);
42             }
43             printf("\n");
44         }
45         printf("!!!\n\n");    */
46         
47         while(q--){
48             scanf("%d%d",&n,&m);
49             int l=min(n,m),s=n+m;
50             if(k==1){
51                 if(l%2){
52                     if(s%2) p1();
53                     else p2();
54                 }
55                 else p1();
56                 continue;
57             }
58             int t=(k+1)*2;
59             if(l%t>=1&&l%t<=k){
60                 if(s%2) p1(); 
61                 else p2();
62             }
63             else if(l%(k+1)==0){
64                 p1();
65             }
66             else{
67                 if(s%2) p2();
68                 else p1();
69             }
70         }
71     }
72     return 0;
73 }

hdu 5708

hdu 5710

容易发现0~4乘2后不损失值，5~9乘2后损失9。设大于5的部分长为l，得到S(n)与l的比例关系，显然约分后数字最小。从低位开始放数字，让低位的尽量大。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int ans[1000000],p,a,b,T,s,l;
 6 
 7 int main(){
 8     scanf("%d",&T);
 9     while(T--){
10         scanf("%d%d",&a,&b);
11         p=0;
12         s=9*b;
13         l=2*b-a;
14         int t=__gcd(s,l);
15         s/=t; l/=t;
16         s=s-l*5;
17         if(l<0||s<0){
18             puts("0");
19             continue;
20         }
21         
22         for(int i=1;i<=l;i++){
23             ans[++p]=5+min(s,4);
24             s=s-min(s,4);
25         }
26         while(s){
27             ans[++p]=min(s,4);
28             s=s-min(s,4);
29         }
30         for(int i=p;i>=1;i--){
31             printf("%d",ans[i]);
32         }
33         printf("\n");
34     }
35     return 0;
36 }

hdu 5710

hdu 6023

水题，模拟。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int T,n,m,r,a,b,ans,cnt,pen[20];
 6 char c;
 7 string s;
 8 bool f[20];
 9 
10 int main(){
11     ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);
12     cin>>T;
13     while(T--){
14         cin>>n>>m;
15         memset(f,0,sizeof(f));
16         memset(pen,0,sizeof(pen));
17         ans=0;    cnt=0;
18         while(m--){
19             cin>>r>>a>>c>>b>>s;
20             r-=1000;
21             if(f[r]) continue;
22             if(s[0]=='A'){
23                 cnt++;
24                 f[r]=1;
25                 ans=ans+a*60+b+pen[r];
26             }
27             else{
28                 pen[r]+=20;
29             }
30         }
31         cout<<cnt<<" "<<ans<<'\n';
32     }
33     return 0;
34 }

hdu 6023

hdu 6024

dp[i][j]为前i个的最小花费，j=1表示第i个建，j=0表示不建。转移如下。其中dp[i][0]需要遍历一遍前面的情况。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 struct node{
 6     ll x,c;
 7     bool operator < (const node &t) const{
 8         return x<t.x;
 9     }
10 }a[3003];
11 
12 ll n,dp[3003][3];
13 
14 int main(){
15     while(~scanf("%lld",&n)){
16         memset(dp,0,sizeof(dp));
17         for(int i=1;i<=n;i++){
18             scanf("%lld%lld",&a[i].x,&a[i].c);
19         }
20         sort(a+1,a+1+n);
21         dp[1][1]=a[1].c;
22         dp[1][0]=(1ll<<60);
23         for(int i=2;i<=n;i++){
24             dp[i][1]=min(dp[i-1][0],dp[i-1][1])+a[i].c;
25             dp[i][0]=(1ll<<60);
26             ll tmp=0;
27             for(int j=i-1;j>=1;j--){
28                 tmp=tmp+(i-j)*(a[j+1].x-a[j].x);
29                 dp[i][0]=min(dp[i][0],dp[j][1]+tmp);
30             }
31         }
32         printf("%lld\n",min(dp[n][0],dp[n][1]));
33     }
34     return 0;
35 }

hdu 6024

hdu 6025

预处理前缀和后缀gcd，根据gcd的性质可以O(n)扫一遍得出答案。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 const int maxn=1e5+20;
 5 
 6 int n,T,a[maxn],pre[maxn],las[maxn],ans;
 7 
 8 int main(){
 9     scanf("%d",&T);
10     while(T--){
11         scanf("%d",&n);
12         for(int i=1;i<=n;i++){
13             scanf("%d",&a[i]);            
14         }
15         pre[1]=a[1]; las[n]=a[n];
16         for(int i=2;i<=n;i++){
17             pre[i]=__gcd(pre[i-1],a[i]);
18         }
19         for(int i=n-1;i>=1;i--){
20             las[i]=__gcd(las[i+1],a[i]);
21         }
22         
23         ans=max(pre[n-1],las[2]);
24         for(int i=2;i<=n-1;i++){
25             ans=max(ans,__gcd(pre[i-1],las[i+1]));
26         }
27         printf("%d\n",ans);
28     }
29     return 0;
30 }

hdu 6025

hdu 6026

稍微修改一下djkstra板子，记录0到每个点的最短路有几条，由于最后一定各保留一条最短路，相乘即为选择的方案数。

 1 #include<bits/stdc++.h>
 2 #define inf 0x3f3f3f3f
 3 #define miaojie1
 4 #ifdef miaojie
 5   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
 6 void err() {std::cout << std::endl;}
 7 template<typename T, typename...Args>
 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
 9 #else
10   #define dbg(...)
11 #endif
12 using namespace std;
13 typedef long long ll;
14 typedef pair<int,int>pii;
15 const ll mod=1e9+7;
16 const int maxn=3003;
17 
18 struct edge{
19     int nxt,to,cost;
20 }e[maxn];
21 
22 int n,head[55],d[55],cnt[55],p;
23 ll ans;
24 char s[55];
25 bool vis[55];
26 priority_queue<pii,vector<pii>,greater<pii> >q;
27 
28 void addedge(int from,int to,int cost){
29     e[p].nxt=head[from];
30     e[p].to=to;
31     e[p].cost=cost;
32     head[from]=p++;
33 }
34 
35 void DJ(){
36     while(!q.empty()) q.pop();
37     for(int i=0;i<n;i++){
38         vis[i]=0;
39         d[i]=inf;
40         cnt[i]=1;
41     }
42     d[0]=0;
43     q.push(make_pair(0,0));    
44     while(!q.empty()){
45         pii t=q.top();
46         dbg(t.first,t.second);
47         q.pop();
48         vis[t.second]=1;
49         for(int i=head[t.second];~i;i=e[i].nxt){
50             if(vis[e[i].to]) continue;
51             if(d[e[i].to]>t.first+e[i].cost){
52                 d[e[i].to]=t.first+e[i].cost;
53                 cnt[e[i].to]=1;
54                 q.push(make_pair(d[e[i].to],e[i].to));
55             }
56             else if(d[e[i].to]==t.first+e[i].cost) cnt[e[i].to]++;
57         }
58     } 
59 }
60 
61 int main(){
62     while(~scanf("%d",&n)){
63         memset(head,-1,sizeof(head)); p=0; ans=1;
64         for(int i=0;i<n;i++){
65             scanf("%s",s);
66             for(int j=0;j<n;j++){
67                 if(s[j]!='0') addedge(i,j,s[j]-48);
68             }
69         }
70         DJ();
71         for(int i=0;i<n;i++){
72             ans=ans*cnt[i]%mod;
73         }
74         printf("%lld\n",ans);
75     }
76     return 0;
77 }

hdu 6026

hdu 6027

水题，预处理一下暴力能过。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 const ll mod=1e9+7;
 6 
 7 int T;
 8 ll n,k,f[10002][7];
 9 
10 void init(){
11     for(int i=1;i<=10000;i++){
12         f[i][0]=1;
13         for(int j=1;j<=5;j++){
14             f[i][j]=f[i][j-1]*i%mod;
15         }
16     }
17 }
18 
19 int main(){
20     scanf("%d",&T);
21     init();
22     while(T--){
23         scanf("%lld%lld",&n,&k);
24         ll ans=0;
25         for(int i=1;i<=n;i++){
26             ans=(ans+f[i][k])%mod;
27         }
28         printf("%lld\n",ans);
29     }
30     return 0;
31 }

hdu 6027

hdu 6029

寻找是否存在完美匹配，从后往前贪心即可，尽量让后面的连接。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int T,n,a[100005]; 
 6 
 7 int main(){
 8     scanf("%d",&T);
 9     while(T--){
10         scanf("%d",&n);
11         int cnt=0;
12         for(int i=2;i<=n;i++){
13             scanf("%d",&a[i]);
14             if(a[i]==1) cnt++;
15         }
16         if(n%2||cnt<n/2){
17             printf("No\n");
18             continue;
19         }
20         
21         int f=0;
22         bool jd=1;
23         for(int i=n;i>=2;i--){
24             if(a[i]==2) f--;
25             else f++;
26             if(f<0){
27                 jd=0;
28                 break;
29             }
30         }
31         if(jd) printf("Yes\n");
32         else printf("No\n");
33     } 
34     return 0;
35 }

hdu 6029

hdu 6030

红记为1，蓝记为0，题目条件转化为连续2个或3个珠子中1的个数>=0的个数。考虑在2个连续的珠子后增加一个，有10-->101->01 , 01->011->11 , 11->111或110->11或10；

记末尾10的个数为f(a)，末尾01的个数为f(b)，末尾11的个数为f(c)，总个数f(n)；

则f(a+1)=f(c)，f(b+1)=f(a)，f(c+1)=f(b)+f(c)；

根据以上关系推出f(n)=f(n-1)+f(n-3)，显然可以用矩阵快速幂解决。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 const int mod=1e9+7;
 6 const int z = 3;
 7 
 8 int T;
 9 ll n;
10 struct ju{
11     long long a[z][z];
12 };
13 
14 ju muli(ju A,ju B){
15     ju C;
16     for(int i=0;i<z;++i){
17         for(int j=0;j<z;++j){
18             C.a[i][j]=0;
19             for(int k=0;k<z;++k){
20                 C.a[i][j]+=A.a[i][k]*B.a[k][j];
21                 C.a[i][j]%=mod;
22             }
23         }
24     }
25     return C;
26 }
27 ju init(){
28     ju res;
29     for(int i=0;i<z;++i){
30         for(int j=0;j<z;++j){
31             if(i==j)res.a[i][j]=1;
32             else res.a[i][j]=0;
33         }
34     }
35     return res;
36 }
37 ju pow(ju A,ll n){
38     ju res=init(),temp=A;
39     for(;n;n/=2){
40         if(n&1)res=muli(res,temp);
41         temp=muli(temp,temp);
42     }
43     return res;
44 }
45 void print(ju A){
46     for(int i=0;i<z;++i){
47         for(int j=0;j<z;++j){
48             cout<<A.a[i][j]<<" ";
49         }
50         cout<<endl;
51     }
52     cout<<endl;
53 }
54 
55 int main(){
56     scanf("%d",&T);
57     while(T--){
58         scanf("%lld",&n);
59         if(n==2) {
60             printf("3\n");
61             continue;
62         }
63         if(n==3){
64             printf("4\n");
65             continue;
66         }
67         ju ans,t;
68         for(int i=0;i<z;i++){
69             for(int j=0;j<z;j++){
70                 ans.a[i][j]=t.a[i][j]=0;
71             }
72         }
73         ans.a[0][0]=6; ans.a[1][0]=4; ans.a[2][0]=3;
74         t.a[0][0]=t.a[0][2]=t.a[1][0]=t.a[2][1]=1;
75     //    print(t);
76         ans=muli(pow(t,n-4),ans);
77     //    print(ans);
78         printf("%lld\n",ans.a[0][0]%mod);
79     }
80     return 0;
81 }

hdu 6030

hdu 6032

在贾老板的指导下勉强混过了。

三个map分别存子串是否出现、出现次数、对应的sg。结构体存胜负，顺便存分数，dfs跑sg的时候记得每次交换自己(p1)和对手(p2)。做着难受，I very vegetable.

 1 #include<bits/stdc++.h>
 2 #define inf 0x3f3f3f3f
 3 #define miaojie1
 4 #ifdef miaojie
 5   #define dbg(args...) do {cout << #args << " : "; err(args);} while (0)
 6 void err() {std::cout << std::endl;}
 7 template<typename T, typename...Args>
 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);}
 9 #else
10   #define dbg(...)
11 #endif
12 using namespace std;
13 
14 struct node{
15     int flag;
16     int p1,p2;
17     node(){}
18     node(bool _flag,int _p1,int _p2) : flag(_flag),p1(_p1),p2(_p2){}
19     bool operator < (const node &a)const{
20         if(flag==a.flag){
21             if(p2==a.p2){
22                 return p1<a.p1;
23             }
24             return p2>a.p2;
25         }
26         return flag<a.flag;
27     }
28 };
29 
30 map<string,int>jd;
31 map<string,set<int> >cnt;
32 map<string,node>sg;
33 int n;
34 char s[100];
35 
36 node dfs(string s){
37     if(sg.count(s)){
38         return sg[s];
39     }
40     node t{1,inf,0};
41     string ts;
42     for(char i='a';i<='z';i++){
43         if(jd.count(s+i))    t=min(t,dfs(s+i));
44         if(jd.count(i+s))    t=min(t,dfs(i+s));
45     }
46     
47     int sum=0,maxi=0;
48     for(int i=0;i<s.size();i++){
49         sum=sum+s[i]-96;
50         maxi=max(maxi,s[i]-96);
51     }
52     sum=sum*maxi+cnt[s].size();
53     dbg(s,t.flag,t.p1,t.p2,sum);
54     if(t.p1==inf){
55         node tt(0,0,sum);
56         return tt;
57     }
58     sg[s].flag=1-t.flag;
59     sg[s].p1=t.p2;
60     sg[s].p2=t.p1+sum;
61     return sg[s];
62 }
63 
64 int main(){
65     while(~scanf("%d",&n)){
66         jd.clear(); cnt.clear(); sg.clear();
67         for(int i=1;i<=n;i++){
68             scanf("%s",&s);
69             int l=strlen(s);
70             for(int j=0;j<l;j++){
71                 string t;
72                 for(int k=j;k<l;k++){
73                     t+=s[k];
74                     jd[t]=1;
75                     cnt[t].insert(i);
76                 }
77             }    
78         }
79         node ans=dfs("");
80         if(ans.flag) printf("Alice\n");
81         else printf("Bob\n");
82         printf("%d %d\n",ans.p1,ans.p2);
83     } 
84     return 0;
85 }

hdu 6032

posted on 2019-06-25 11:54 喵杰阅读(1331) 评论(0) 编辑收藏举报