2017 ACM/ICPC 沈阳 F题 Heron and his triangle

A triangle is a Heron’s triangle if it satisfies that the side lengths of it are consecutive integers t−1, t, t+ 1 and thatits area is an integer. Now, for given n you need to find a Heron’s triangle associated with the smallest t bigger 
than or equal to n.

InputThe input contains multiple test cases. The first line of a multiple input is an integer T (1 ≤ T ≤ 30000) followedby T lines. Each line contains an integer N (1 ≤ N ≤ 10^30). 
OutputFor each test case, output the smallest t in a line. If the Heron’s triangle required does not exist, output -1.Sample Input

4
1
2
3
4

Sample Output

4
4
4
4
题解：大数.找规律.a[n]=4*a[n-1]-a[n-2]（附上C++大数模板）
  1 #include <bits/stdc++.h>
  2 #define rep(i,a,n) for(int i=a;i<n;++i)
  3 #define per(i,a,n) for(int i=n-1;i>=a;--i)
  4 #define fi first
  5 #define se second
  6 using namespace std;
  7 // base and base_digits must be consistent
  8 constexpr int base = 1000000000;
  9 constexpr int base_digits = 9;
 10 struct bigint{
 11     vector<int> z;
 12     int sign;
 13     bigint() : sign(1) {}
 14     bigint(long long v) { *this = v; }
 15     bigint& operator=(long long v)
 16     {
 17         sign = v < 0 ? -1 : 1;
 18         v*=sign;
 19         z.clear();
 20         for(; v > 0; v = v / base) z.push_back((int)(v % base));
 21         return *this;
 22     }
 23 
 24     bigint(const string& s) { read(s); }
 25 
 26     bigint& operator+=(const bigint& other)
 27     {
 28         if (sign == other.sign)
 29         {
 30             for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
 31             {
 32                 if(i==z.size()) z.push_back(0);
 33                 z[i] += carry + (i < other.z.size() ? other.z[i] : 0);
 34                 carry = z[i] >= base;
 35                 if(carry) z[i] -= base;
 36             }
 37         }
 38         else if (other != 0 /* prevent infinite loop */)
 39         {
 40             *this -= -other;
 41         }
 42         return *this;
 43     }
 44 
 45     friend bigint operator+(bigint a, const bigint& b)
 46     {
 47         return a += b;
 48     }
 49 
 50     bigint& operator-=(const bigint& other)
 51     {
 52         if (sign == other.sign)
 53         {
 54             if (sign == 1 && *this >= other || sign == -1 && *this <= other)
 55             {
 56                 for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
 57                 {
 58                     z[i] -= carry + (i < other.z.size() ? other.z[i] : 0);
 59                     carry = z[i] < 0;
 60                     if(carry) z[i] += base;
 61                 }
 62                 trim();
 63             }
 64             else
 65             {
 66                 *this = other - *this;
 67                 this->sign = -this->sign;
 68             }
 69         }
 70         else *this += -other;
 71         return *this;
 72     }
 73 
 74     friend bigint operator - (bigint a,const bigint& b)
 75     {
 76         return a -= b;
 77     }
 78 
 79     bigint& operator*=(int v)
 80     {
 81         if(v<0) sign=-sign,v=-v;
 82         for(int i=0,carry=0;i<z.size() || carry;++i)
 83         {
 84             if(i==z.size()) z.push_back(0);
 85             long long cur = (long long)z[i] * v + carry;
 86             carry = (int)(cur / base);
 87             z[i] = (int)(cur % base);
 88         }
 89         trim();
 90         return *this;
 91     }
 92 
 93     bigint operator*(int v) const
 94     {
 95         return bigint(*this) *= v;
 96     }
 97 
 98     friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1)
 99     {
100         int norm = base / (b1.z.back() + 1);
101         bigint a = a1.abs() * norm;
102         bigint b = b1.abs() * norm;
103         bigint q, r;
104         q.z.resize(a.z.size());
105 
106         for (int i = (int)a.z.size() - 1; i >= 0; i--)
107         {
108             r*=base; r+=a.z[i];
109             int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
110             int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
111             int d = (int)(((long long)s1 * base + s2) / b.z.back());
112             r -= b * d;
113             while(r < 0) r+=b,--d;
114             q.z[i] = d;
115         }
116 
117         q.sign = a1.sign * b1.sign;
118         r.sign = a1.sign;
119         q.trim();
120         r.trim();
121         return {q, r / norm};
122     }
123 
124     friend bigint sqrt(const bigint& a1)
125     {
126         bigint a=a1;
127         while(a.z.empty()||a.z.size()%2==1) a.z.push_back(0);
128 
129         int n = a.z.size();
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) a.z.push_back(0);
135 
136         bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
137         firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
138         int q = firstDigit;
139         bigint res;
140         for (int j = n / 2 - 1; j >= 0; j--)
141         {
142             for(;;--q)
143             {
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) { r=r1; break; }
146             }
147             res*=base;res+=q;
148             if(j>0)
149             {
150                 int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
151                 int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
152                 int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()]:0;
153                 q = (int)(((long long)d1*base*base+(long long)d2*base+d3)/(firstDigit*2));
154             }
155         }
156 
157         res.trim();
158         return res / norm;
159     }
160 
161     bigint operator/(const bigint& v) const
162     {
163         return divmod(*this, v).first;
164     }
165 
166     bigint operator%(const bigint& v) const
167     {
168         return divmod(*this, v).second;
169     }
170 
171     bigint& operator/=(int v)
172     {
173         if(v<0) sign=-sign,v=-v;
174         for (int i = (int)z.size() - 1, rem = 0; i >= 0; --i)
175         {
176             long long cur = z[i] + rem * (long long)base;
177             z[i] = (int)(cur / v);
178             rem = (int)(cur % v);
179         }
180         trim();
181         return *this;
182     }
183 
184     bigint operator/(int v) const
185     {
186         return bigint(*this) /= v;
187     }
188 
189     int operator%(int v) const
190     {
191         if(v<0) v=-v;
192         int m=0;
193         for(int i=(int)z.size()-1;i>=0;--i) m=(int)((z[i]+m*(long long)base)%v);
194         return m * sign;
195     }
196 
197     bigint& operator*=(const bigint& v)
198     {
199         *this = *this * v;
200         return *this;
201     }
202 
203     bigint& operator/=(const bigint& v)
204     {
205         *this = *this / v;
206         return *this;
207     }
208 
209     bool operator<(const bigint& v) const
210     {
211         if(sign!=v.sign) return sign < v.sign;
212         if(z.size()!=v.z.size()) return z.size()*sign<v.z.size()*v.sign;
213         for(int i = (int)z.size() - 1; i >= 0; i--)
214             if(z[i] != v.z[i])  return z[i] * sign < v.z[i] * sign;
215         return false;
216     }
217 
218     bool operator>(const bigint& v) const { return v < *this; }
219     bool operator<=(const bigint& v) const { return !(v < *this); }
220     bool operator>=(const bigint& v) const { return !(*this < v); }
221     bool operator==(const bigint& v) const { return !(*this < v) && !(v < *this); }
222     bool operator!=(const bigint& v) const { return *this < v || v < *this; }
223 
224     void trim()
225     {
226         while(!z.empty() && z.back() == 0) z.pop_back();
227         if(z.empty()) sign = 1;
228     }
229 
230     bool isZero() const { return z.empty(); }
231 
232     friend bigint operator-(bigint v)
233     {
234         if(!v.z.empty()) v.sign = -v.sign;
235         return v;
236     }
237 
238     bigint abs() const
239     {
240         return sign == 1 ? *this : -*this;
241     }
242 
243     long long longValue() const
244     {
245         long long res = 0;
246         for(int i = (int)z.size() - 1; i >= 0; i--) res = res * base + z[i];
247         return res * sign;
248     }
249 
250     friend bigint gcd(const bigint& a, const bigint& b)
251     {
252         return b.isZero() ? a : gcd(b, a % b);
253     }
254 
255     friend bigint lcm(const bigint& a, const bigint& b)
256     {
257         return a / gcd(a, b) * b;
258     }
259 
260     void read(const string& s)
261     {
262         sign = 1;
263         z.clear();
264         int pos = 0;
265         while(pos < s.size() && (s[pos] == '-' || s[pos] == '+'))
266         {
267             if(s[pos] == '-') sign = -sign;
268             ++pos;
269         }
270         for(int i=(int)s.size()-1;i>=pos;i-=base_digits)
271         {
272             int x=0;
273             for(int j=max(pos,i-base_digits+1);j<=i;j++) x=x*10+s[j]-'0';
274             z.push_back(x);
275         }
276         trim();
277     }
278 
279     friend istream& operator>>(istream& stream, bigint& v)
280     {
281         string s;
282         stream >> s;
283         v.read(s);
284         return stream;
285     }
286 
287     friend ostream& operator<<(ostream& stream, const bigint& v)
288     {
289         if(v.sign == -1) stream << '-';
290         stream << (v.z.empty() ? 0 : v.z.back());
291         for(int i = (int)v.z.size() - 2; i >= 0; --i)
292             stream << setw(base_digits) << setfill('0') << v.z[i];
293         return stream;
294     }
295 
296     static vector<int> convert_base(const vector<int>& a, int old_digits, int new_digits)
297     {
298         vector<long long> p(max(old_digits, new_digits) + 1);
299         p[0] = 1;
300         for(int i=1;i<p.size();i++) p[i]=p[i-1]*10;
301         vector<int> res;
302         long long cur = 0;
303         int cur_digits = 0;
304         for(int v : a)
305         {
306             cur += v * p[cur_digits];
307             cur_digits += old_digits;
308             while (cur_digits >= new_digits)
309             {
310                 res.push_back(int(cur % p[new_digits]));
311                 cur /= p[new_digits];
312                 cur_digits -= new_digits;
313             }
314         }
315         res.push_back((int)cur);
316         while(!res.empty() && res.back()==0)
317             res.pop_back();
318         return res;
319     }
320 
321     typedef vector<long long> vll;
322     static vll karatsubaMultiply(const vll& a, const vll& b)
323     {
324         int n=a.size();
325         vll res(n + n);
326         if(n <= 32)
327         {
328             for (int i = 0; i < n; i++)
329                 for (int j = 0; j < n; j++)
330                     res[i + j] += a[i] * b[j];
331             return res;
332         }
333 
334         int k = n >> 1;
335         vll a1(a.begin(), a.begin() + k);
336         vll a2(a.begin() + k, a.end());
337         vll b1(b.begin(), b.begin() + k);
338         vll b2(b.begin() + k, b.end());
339         vll a1b1 = karatsubaMultiply(a1, b1);
340         vll a2b2 = karatsubaMultiply(a2, b2);
341         for(int i=0;i<k;i++) a2[i]+=a1[i];
342         for(int i=0;i<k;i++) b2[i]+=b1[i];
343 
344         vll r = karatsubaMultiply(a2, b2);
345         for(int i=0;i<a1b1.size();i++) r[i]-=a1b1[i];
346         for(int i=0;i<a2b2.size();i++) r[i]-=a2b2[i];
347         for(int i=0;i<r.size();i++) res[i+k]+=r[i];
348         for(int i=0;i<a1b1.size();i++) res[i]+=a1b1[i];
349         for(int i = 0;i<a2b2.size();i++) res[i+n]+=a2b2[i];
350         return res;
351     }
352 
353     bigint operator*(const bigint& v) const
354     {
355         vector<int> a6=convert_base(this->z,base_digits,6);
356         vector<int> b6=convert_base(v.z,base_digits,6);
357         vll a(a6.begin(),a6.end());
358         vll b(b6.begin(),b6.end());
359         while(a.size()<b.size()) a.push_back(0);
360         while(b.size()<a.size()) b.push_back(0);
361         while(a.size()&(a.size()-1)) a.push_back(0),b.push_back(0);
362         vll c=karatsubaMultiply(a, b);
363         bigint res;
364         res.sign = sign * v.sign;
365         for (int i = 0, carry = 0; i < c.size(); i++)
366         {
367             long long cur = c[i] + carry;
368             res.z.push_back((int)(cur % 1000000));
369             carry = (int)(cur / 1000000);
370         }
371         res.z = convert_base(res.z, 6, base_digits);
372         res.trim();
373         return res;
374     }
375 };
376 
377 bigint qpow(bigint a,bigint b){
378     bigint ans=1;
379     while(b!=0){
380         if(b%2){
381             ans= ans*a;
382         }
383         b/=2;
384         a= a*a;
385     }
386     return ans;
387 
388 }
389 
390 
391 struct Matrix
392 {
393     bigint a[2][2];
394     Matrix()
395     {
396         rep(i,0,2){
397             rep(j,0,2){
398                 a[i][j]=0;
399             }
400         }
401     }
402     Matrix operator * (const Matrix y)
403     {
404         Matrix ans;
405         for(int i = 0; i < 2; i++)
406             for(int j = 0; j < 2; j++)
407                 for(int k = 0; k < 2; k++)
408                     ans.a[i][j] = ans.a[i][j] + (a[i][k]*y.a[k][j]);
409         return ans;
410     }
411     Matrix operator = (const Matrix y)
412     {
413         for(int i=0;i<2;i++)
414             for(int j=0;j<2;j++)
415                 a[i][j]=y.a[i][j];
416     }
417     Matrix operator *= (const Matrix y)
418     {
419         Matrix ans;
420         for(int i=0;i<2;i++)
421             for(int j=0;j<2;j++)
422                 for(int k=0;k<2;k++)
423                     ans.a[i][j] = ans.a[i][j] + (a[i][k]*y.a[k][j]);
424 
425         for(int i=0;i<2;i++)
426             for(int j=0;j<2;j++)
427                 a[i][j]=ans.a[i][j];
428     }
429 };
430 
431 Matrix qpow(bigint x)
432 {
433     Matrix ans;
434     ans.a[0][0]=ans.a[1][1]=1; //单位矩阵
435     Matrix mul;
436     mul.a[0][0]=4;
437     mul.a[0][1]=-1;
438     mul.a[1][0]=1;
439     mul.a[1][1]=0;
440     while(x!=0)
441     {
442         if(x%2!=0)
443             ans = ans*mul;
444         mul = mul* mul;
445         x/=2;
446     }
447     return ans;
448 }
449 bigint ans[1005];
450 void solve(){
451 
452 
453     ans[0]=(bigint)4;
454     ans[1]=(bigint)14;
455     ans[2]=(bigint)52;
456     rep(i,2,200){
457         ans[i]=(bigint)4*ans[i-1]-ans[i-2];
458 //        cout<<ans[i]<<endl;
459     }
460 
461 }
462 int main()
463 {
464     solve();
465     int t;cin>>t;
466     while(t--){
467         bigint n;
468         cin>>n;
469         rep(i,0,200){
470             if(ans[i]>=n){
471                 cout<<ans[i]<<endl;
472                 break;
473             }
474         }
475 
476 
477     }
478 }
View Code

 

　　

posted @ 2018-10-14 01:07  StarHai  阅读(463)  评论(0)  编辑  收藏  举报