17青岛网络赛

接连爆炸...

 

A Cubic number and A Cubic Number

 HDU - 6216 

题意:给素数p,问是否存在两个数的立方差等于x.

a3-b3立方差公式展开得(a-b)*(a2+ab+b2) == p 

p是素数,所以a-b等于1,于是可以得到3a2+3a+1 == p  即 a(a+1) == (p-1)/3

---场上我没看,队友写的...

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long 
 4 
 5 int main(){
 6     int t;
 7     scanf("%d",&t);
 8     ll x;
 9     while(t--){
10         scanf("%lld",&x);
11         x-=1;
12         if(x%3){
13             puts("NO");
14         }else{
15           x=x/3;
16           ll a=sqrt(x);
17           if(a*(a+1)==x) puts("YES");
18           else puts("NO");
19         }
20     }
21     return 0;
22 }
View Code

 

Chinese Zodiac

 HDU - 6213 

题意:给生肖,算年龄差~

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 map<string,int> mp;
 5 
 6 int main(){
 7     int t;
 8     scanf("%d",&t);
 9     mp["rat"]=1; mp["ox"]=2; mp["tiger"]=3; mp["rabbit"]=4;
10     mp["dragon"]=5; mp["snake"]=6; mp["horse"]=7; mp["sheep"]=8;
11     mp["monkey"]=9; mp["rooster"]=10; mp["dog"]=11; mp["pig"]=12;
12     while(t--){
13         string a,b;
14         cin>>a>>b;
15         int u=mp[a], v=mp[b];
16         int ans=(v-u+11)%12+1;
17         cout<<ans<<endl;
18     }
19 }
View Code

 

The Dominator of Strings

 HDU - 6208 

题意:给n个字符串,问其中是否存在一个串包含其他所有串.

好像方法很多...

场上一开始想用ac自动机搞,然后不知道为什么把自己否了=_=,,

队友用后缀自动机搞一直TLE,我没学过也帮不上忙...

  1 /*************************************************************************
  2     > File Name: a.cpp
  3     > Author: yijiull
  4     > Mail: 1147161372@qq.com 
  5     > Create
  6     Time: 2017年09月18日 星期一 16时00分57秒
  7  ************************************************************************/
  8 
  9 #include<iostream>
 10 #include<cstring>
 11 #include<cstdio>
 12 #include <bits/stdc++.h>
 13 using namespace std;
 14 const int maxnode = 100010;
 15 const int sigma = 26;
 16 
 17 struct AC{
 18     int ch[maxnode][sigma], last[maxnode], f[maxnode];
 19     int val[maxnode];
 20     int sz;
 21     int cnt;
 22 
 23     void init(){
 24         sz = 1;
 25         cnt = 0;
 26         val[0] = 0;
 27         memset(ch[0],0,sizeof(ch[0]));
 28     }
 29 
 30     int idx(char s) {
 31         return s - 'a';
 32     }
 33 
 34     void insert_(char *s){
 35         int u = 0, n = strlen(s);
 36         for(int i = 0; i < n; i++){
 37             int c = idx(s[i]);
 38             if(!ch[u][c]){
 39                 memset(ch[sz],0,sizeof(ch[sz]));
 40                 val[sz] = 0;
 41                 ch[u][c] = sz++;
 42             }
 43             u = ch[u][c];
 44         }
 45         val[u]++;
 46     }
 47 
 48     void getfail(){
 49         queue<int> q;
 50         f[0] = 0;
 51         for(int c = 0; c < sigma; c++){
 52             int u = ch[0][c];
 53             if(u){
 54                 f[u] = 0;
 55                 q.push(u);
 56                 last[u] = 0;
 57             }
 58         }
 59         while(!q.empty()){
 60             int r = q.front();
 61             q.pop();
 62             for(int c = 0; c < sigma; c++){
 63                 int u = ch[r][c];
 64                 if(!u){
 65                     ch[r][c] = ch[f[r]][c];
 66                     continue;
 67                 }
 68                 q.push(u);
 69                 int v = f[r];
 70                 while(v && !ch[v][c]) v = f[v];
 71                 f[u] = ch[v][c];
 72                 last[u] = val[f[u]] ? f[u] : last[f[u]]; 
 73             }
 74         }
 75     }
 76     
 77     void print(int u){
 78         if(u){
 79             cnt += val[u];
 80             val[u] = 0;
 81             print(last[u]);
 82         }
 83     }
 84     
 85     void query(char *s){
 86         int u = 0, n = strlen(s);
 87         for(int i = 0; i < n; i++){
 88             int c=idx(s[i]);
 89             while(u && !ch[u][c]) u = f[u];
 90             u = ch[u][c];
 91             int temp = 0;
 92             if(val[u]) temp = u;
 93             else if(last[u]) temp = last[u];
 94             while(temp){
 95                 cnt += val[temp];
 96                 val[temp] = 0;
 97                 temp = last[temp];
 98             }
 99         }
100     }
101 
102 };
103 AC ac;
104 char s[maxnode];
105 char ans[maxnode];
106 
107 int main(){
108     int t;
109     scanf("%d", &t);
110     while(t--){
111         ac.init();
112         int n;
113         scanf("%d", &n);
114         int maxl = 0;
115         for(int i = 0; i < n; i++){
116             scanf("%s",s);
117             int len = strlen(s);
118             if(maxl < len){
119                 maxl = len;
120                 strcpy(ans,s);
121             }
122             ac.insert_(s);
123         }
124         ac.getfail();
125         ac.query(ans);
126         if(ac.cnt == n){
127             puts(ans);
128         }else{
129             puts("No");
130         }
131     }
132     return 0;
133 }
ac自动机

 

看到别人直接KMP更快...

strcpy(char *p, char *s) ; //把s复制给p, 包括最后的'\0'

strncpy(char *p, char *s, int n) ;  // 把s的前n个字符复制给p, 不加'\0'

<  strncpy并没有拷贝串后的\0字符，而strcpy却拷贝了。

<  这充分说明，strncpy是为拷贝字符而生的，而strcpy是拷贝字符串而生的。但两者都不能越界拷贝。只要正确使用strncpy, 那就比strcpy安全。

然而我写的KMP还是和ac自动机一样的慢...复制字符串太耗费时间=_=

 1 /*************************************************************************
 2     > File Name: a.cpp
 3     > Author: yijiull
 4     > Mail: 1147161372@qq.com 
 5     > Created Time: 2017年09月18日 星期一 17时22分46秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include <bits/stdc++.h>
12 using namespace std;
13 const int maxn=100010;
14 char s[maxn], p[maxn], str[maxn];
15 int f[maxn];
16 
17 int a[maxn],b[maxn];
18 int lenp, lens;
19 void getfail(char *p){
20     f[0] = f[1] = 0;
21     for(int i = 1; i < lenp; i++){
22         int j = f[i];
23         while(j && p[j]!=p[i]) j = f[j];
24         f[i+1] = p[i]==p[j] ? j+1 : 0;
25     }
26     return ;
27 }
28 int kmp(char *s, char *p){
29     lenp = strlen(p);
30     lens = strlen(s);
31     getfail(p);
32     int j = 0;
33     for(int i = 0; i < lens; i++){
34         while(j &&s[i]!=p[j]) j = f[j];
35         if(s[i] == p[j]) j++;
36         if(j == lenp) return 1;
37     }
38     return 0;
39 }
40 
41 int main(){
42     int t;
43     scanf("%d",&t);
44     while(t--){
45         int n;
46         int len = 0;
47         int maxlen = 0;
48         scanf("%d", &n);
49         for(int i = 0; i < n; i++) {
50             scanf("%s", str+len);
51             a[i] = len;
52             b[i] = strlen(str);
53             len += b[i] - a[i];
54             if(b[i] - a[i] > maxlen) {
55                 maxlen = b[i] - a[i];
56                 strncpy(s,&str[a[i]],b[i]-a[i]);
57                 s[b[i]-a[i]]=0;
58             }
59         }
60         int flag = 1;
61         for(int i = 0; i < n; i++) {
62             strncpy(p,&str[a[i]],b[i]-a[i]);
63             p[b[i]-a[i]]=0;
64             if(!kmp(s,p)) {
65                 flag = 0;
66                 break;
67             }
68         }
69         if(flag) puts(s);
70         else puts("No");
71     }
72     return 0;
73 }
KMP

 

下面这个是参考的大佬的~ 其实字符串不用复制 ,整整快了2s......

 1 /*************************************************************************
 2     > File Name: a.cpp
 3     > Author: yijiull
 4     > Mail: 1147161372@qq.com 
 5     > Created Time: 2017年09月18日 星期一 17时22分46秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include <bits/stdc++.h>
12 using namespace std;
13 const int maxn=100010;
14 char str[maxn*100];
15 int f[maxn];
16 char *p[maxn];
17 char *s;
18 int lenp, lens;
19 
20 void getfail(char *p){
21     f[0] = f[1] = 0;
22     for(int i = 1; i < lenp; i++){
23         int j = f[i];
24         while(j && p[j]!=p[i]) j = f[j];
25         f[i+1] = p[i]==p[j] ? j+1 : 0;
26     }
27     return ;
28 }
29 int kmp(char *s, char *p){
30     lenp = strlen(p);
31     lens = strlen(s);
32     getfail(p);
33     int j = 0;
34     for(int i = 0; i < lens; i++){
35         while(j &&s[i]!=p[j]) j = f[j];
36         if(s[i] == p[j]) j++;
37         if(j == lenp) return 1;
38     }
39     return 0;
40 }
41 
42 int main(){
43     int t;
44     scanf("%d",&t);
45     while(t--){
46         int n;
47         int len = 0;
48         int maxlen = 0;
49         scanf("%d", &n);
50         char *o=str;
51         for(int i = 0; i < n; i++) {
52             scanf("%s", o);
53             len = strlen(o);
54             if(len > maxlen) {
55                 maxlen = len;
56                 s = o;
57             }
58             p[i] = o;
59             o += strlen(o) + 2;  //!!!
60         }
61 
62         int flag = 1;
63         for(int i = 0; i < n; i++) {
64             if(!kmp(s, p[i])) {
65                 flag = 0;
66                 break;
67             }
68         }
69         if(flag) puts(s);
70         else puts("No");
71     }
72     return 0;
73 }
KMP

 

还可以直接用string的find函数过......

还可以用java过,而且很快......

醉了~

 

 

Apple

 HDU - 6206 

题意:给四个点,前三个点确定一个外接圆,判断第四个点是否在圆外.

我看的这个题...一开始没注意到精度写了个c++,写完发现精度不够...

java还不会=_= 

 1 /*************************************************************************
 2     > File Name: Main.java
 3     > Author: yijiull
 4     > Mail: 1147161372@qq.com 
 5     > Created Time: 2017年09月21日 星期四 19时26分43秒
 6  ************************************************************************/
 7 import java.math.BigDecimal;
 8 import java.util.*;
 9 import java.io.*;
10 public class Main{
11     public static void main(String[] args){
12         Scanner cin = new Scanner(new BufferedInputStream(System.in));
13         BigDecimal x1, x2, y1, y2, x3, y3, x4, y4;
14         int n;
15         n = cin.nextInt();
16         while(n-- != 0){
17             x1 = cin.nextBigDecimal();
18             y1 = cin.nextBigDecimal();
19             x2 = cin.nextBigDecimal();
20             y2 = cin.nextBigDecimal();
21             x3 = cin.nextBigDecimal();
22             y3 = cin.nextBigDecimal();
23             x4 = cin.nextBigDecimal();
24             y4 = cin.nextBigDecimal();
25 
26             BigDecimal v1 =(y1.multiply(y1).add(x1.multiply(x1))).subtract(x3.multiply(x3)).subtract(y3.multiply(y3));
27             BigDecimal v2 =(y1.multiply(y1).add(x1.multiply(x1))).subtract(x2.multiply(x2)).subtract(y2.multiply(y2));
28             BigDecimal t1 = x1.subtract(x2);
29             BigDecimal t2 = x1.subtract(x3);
30             BigDecimal t3 = y1.subtract(y3);
31             BigDecimal t4 = y1.subtract(y2);
32             BigDecimal y0 = (v1.multiply(t1).subtract(v2.multiply(t2))).divide(t1.multiply(t3).subtract(t2.multiply(t4)));
33             y0 = y0.divide(BigDecimal.valueOf(2));
34 
35             t1 = y1.subtract(y2);
36             t2 = y1.subtract(y3);
37             t3 = x1.subtract(x3);
38             t4 = x1.subtract(x2);
39             BigDecimal x0 = (v1.multiply(t1).subtract(v2.multiply(t2))).divide(t1.multiply(t3).subtract(t2.multiply(t4)));
40             x0 = x0.divide(BigDecimal.valueOf(2));
41     
42             //System.out.println(x0);
43             //System.out.println(y0);
44             
45             BigDecimal r = x0.subtract(x1).multiply(x0.subtract(x1)).add(y0.subtract(y1).multiply(y0.subtract(y1)));
46             BigDecimal d = x0.subtract(x4).multiply(x0.subtract(x4)).add(y0.subtract(y4).multiply(y0.subtract(y4)));
47             
48             if(d.compareTo(r) > 0){
49                 System.out.println("Accepted");
50             } else{
51                 System.out.println("Rejected");
52             }
53         }
54     }
55 }
View Code

算是第一道用java做的题了~

 

 

Smallest Minimum Cut

 HDU - 6214

题意：求最小割的最少边数．

建图边权为　w*(m+1)+1,  求出最大流模(m+1)就是答案．最大流除以(m+1)就是原图最大流．

 

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 const int inf = 0x3f3f3f3f;
  4 const int maxv = 210;
  5 const int maxe = 1010;
  6 
  7 struct Edge{
  8     int u, v, nex;
  9     int flow, cap;
 10     Edge(int u=0, int v=0, int nex=0, int flow=0, int cap=0):
 11         u(u), v(v), nex(nex), flow(flow), cap(cap){}
 12 }e[maxe<<1];
 13 int head[maxv];
 14 int cnt;
 15 void init(){
 16     memset(head, -1, sizeof(head));
 17     cnt = 0;
 18 }
 19 void addEdge(int u, int v, int cap){
 20     e[cnt] = Edge(u, v, head[u], 0, cap);
 21     head[u] = cnt++;
 22     e[cnt] = Edge(v, u, head[v], 0, 0);
 23     head[v] = cnt++;
 24 }
 25 
 26 int S, T, N;
 27 int vis[maxv], d[maxv], num[maxv], cur[maxv], p[maxv];
 28 
 29 void bfs(){
 30     queue<int> q;
 31     q.push(T);
 32     memset(vis, 0, sizeof(vis));
 33     memset(d, -1, sizeof(d));
 34     vis[T] = 1;
 35     d[T] = 0;
 36     while(!q.empty()){
 37         int u = q.front();
 38         q.pop();
 39         for(int i = head[u]; ~i; i=e[i].nex){
 40             int id = i&(-2);
 41             int v = e[id].u;
 42             if(!vis[v] && e[id].cap > e[id].flow){
 43                 d[v] = d[u] + 1;
 44                 vis[v] = 1;
 45                 q.push(v);
 46             }
 47         }
 48     }
 49 }
 50 int augment(){
 51     int u = T, a = inf;
 52     while(u != S){
 53         int id = p[u];
 54         a = min(a, e[id].cap - e[id].flow);
 55         u = e[id].u;
 56     }
 57     u = T;
 58     while(u != S){
 59         int id = p[u];
 60         e[id].flow += a;
 61         e[id^1].flow -= a;
 62         u = e[id].u;
 63     }
 64     return a;
 65 }
 66 
 67 int ISAP(){
 68     bfs();
 69     int flow = 0;
 70     memset(num, 0, sizeof(num));
 71     for(int i = 0; i < N; i++){
 72         cur[i] = head[i];
 73         if(~d[i]) num[d[i]]++;
 74     }
 75     int u = S;
 76     while(d[S] < N){
 77         if(u == T){
 78             flow += augment();
 79             u = S;
 80         }
 81         int ok = 0;
 82         for(int i = cur[u]; ~i; i=e[i].nex){
 83             int v = e[i].v;
 84             if(d[u] == d[v]+1 && e[i].cap > e[i].flow){
 85                 p[v] = i;
 86                 ok = 1;
 87                 cur[u] = i;
 88                 u = v;
 89                 break;
 90             }
 91         }
 92         if(!ok){
 93             int m = N-1;
 94             for(int i = head[u]; ~i; i=e[i].nex){
 95                 if(e[i].cap > e[i].flow && ~d[e[i].v]) m = min(m, d[e[i].v]);
 96             }
 97             if(--num[d[u]] == 0) break;
 98             num[d[u] = m+1]++;
 99             cur[u] = head[u];
100             if(u != S) u = e[p[u]].u;
101         }
102     }
103     return flow;
104 }
105 
106 int main(){
107     int n, m, t;
108     //freopen("in.txt", "r", stdin);
109     scanf("%d", &t);
110     while(t--){
111         init();
112         scanf("%d %d", &n, &m);
113         scanf("%d %d", &S, &T);
114         S--; T--;
115         for(int i = 0; i < m; i++){
116             int u, v, cap;
117             scanf("%d %d %d", &u, &v, &cap);
118             u--; v--;
119             addEdge(u, v, cap*(m+1)+1);
120         }
121 
122         N = n;
123         int ans = ISAP();
124         printf("%d\n", ans%(m+1));
125     }
126 }
View Code

 

posted @ 2017-09-17 18:03  yijiull  阅读(239)  评论(0)  编辑  收藏  举报