The 11th Zhejiang University Programming Contest C G J

来源：互联网发布：淘宝卖家如何删微淘编辑：程序博客网时间：2024/04/30 13:20

Chinese Zodiac

Gaussian Prime

Judge Internal Error

只贴代码，不解释。。T T 。。。

I题木有搞出来。。。哎。。。

Final Rank 99 T T 。。

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;char str[13][15] = {"","Rat", "Ox", "Tiger", "Rabbit", "Dragon", "Snake", "Horse", "Ram", "Monkey", "Rooster", "Dog","Pig"};int main(){int ncases,n;scanf("%d",&ncases);while( ncases-- ){scanf("%d",&n);printf("%s/n",str[12-(n+7)%12]);}return 0;}

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;const int MAX = 50000;int prim[MAX];void init(){fill(prim,prim+MAX,1);prim[0] = prim[1] = 0;for(int i=2; i<MAX; i++)for(int k=2; k*i<MAX; k++)if( prim[i] == 1 )prim[k*i] = 0;for(int n=0; 4*n+3<MAX; n++)if( prim[4*n+3] == 1 )prim[4*n+3] = 2;for(int i=1; i*i<MAX; i++)for(int k=1; k*k+i*i<MAX; k++)if( prim[i*i+k*k] == 1 )prim[i*i+k*k] = 3;}int gcd(int n,int m){return m == 0 ? n : gcd(m,n%m);}int main(){int ncases;int x1,x2,y1,y2;init();scanf("%d",&ncases);while( ncases-- ){scanf("%d %d %d %d",&x1,&x2,&y1,&y2);int num = ((x2-x1)+1)*((y2-y1)+1);int count = 0;for(int i=x1; i<=x2; i++)for(int k=y1; k<=y2; k++){if( i == 0 && k != 0 )if( k > 0 && prim[k] == 2 || k < 0 && prim[-k] == 2 )count++;if( k == 0 && i != 0 )if( i > 0 && prim[i] == 2 || i < 0 && prim[-i] == 2 )count++;if( i != 0 && k != 0 && prim[i*i+k*k] == 3) count++;}int wuyu = gcd(count,num);printf("%d/%d/n",count/wuyu,num/wuyu);}return 0;}

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;int a[5000];int main(){int ncases,n;scanf("%d",&ncases);while( ncases-- ){scanf("%d",&n);for(int i=0; i<n; i++)scanf("%d",&a[i]);sort(a,a+n);int max = 0,count = 0;int tmp = a[0],now = a[0];a[n] = -1;for(int i=1; i<=n; i++){if( a[i] == tmp )count++;else{if( count >= max ){max = count;now = tmp;}count = 0;tmp = a[i];}}printf("%d/n",now);}return 0;}