Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 3574    Accepted Submission(s): 1401

Problem Description

Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
  Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.

 

Input

The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). A_i,j = 1 means i-th people loves j-th people, otherwise A_i,j = 0.
It is guaranteed that the given relationship is a tournament, that is, A_i,i= 0, A_i,j ≠ A_j,i(1<=i, j<=n,i≠j).

 

Output

For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”.
Take the sample output for more details.

 

Sample Input

25001001000001001111011100050111100000010000110001110

 

Sample Output

Case #1: YesCase #2: No

#include<stdio.h>#include<string.h>char p[2001][2001];int s[2001],m;void TO(int n){int flag=0;int i,j,k;for(i=0;i<n;i++){for(j=0;j<n;j++)if(s[j]==0)break;if(j==n){flag=1;break;}s[j]=-1;for(k=0;k<n;k++)if(p[j][k]=='1'&&s[k]!=0)s[k]--; }printf("Case #%d: ",++m);if(flag) printf("Yes\n"); else printf("No\n");}int main(){int t,n;m=0;scanf("%d",&t);while(t--){memset(s,0,sizeof(s));scanf("%d",&n);for(int i=0;i<n;i++){scanf("%s",p[i]);for(int j=0;j<n;j++){if(p[i][j]=='1')s[j]++;}}TO(n);}return 0;}

再贴一个代码：

#include<stdio.h>#include<string.h>char p[2001][2001];int s[2001],m;void TO(int n){int flag=0,i,j,k,w=1000000,h;for(i=0;i<n;i++){for(j=0;j<n;j++)if(s[j]==0){w=j;break;}else w=1000000;if(w==1000000){printf("Case #%d: ",++m);printf("Yes\n");return ;     }s[w]=-1;for(k=0;k<n;k++)if(p[w][k]=='1')s[k]--; }printf("Case #%d: ",++m);printf("No\n");return ;}int main(){int t,n;m=0;scanf("%d",&t);while(t--){memset(s,0,sizeof(s));scanf("%d",&n);for(int i=0;i<n;i++){scanf("%s",p[i]);for(int j=0;j<n;j++){if(p[i][j]=='1')s[j]++;}}TO(n);}return 0;}

问题 B: 此题乃神题，劝你别做

时间限制: 1 Sec  内存限制: 128 MB
提交: 137  解决: 8
[提交][状态][讨论版]

题目描述

声明：这道题没有涉及任何算法！给定函数f = (1) + (2) * b + (3) * c + (4) * d + (5)。

输入

输入数据有多组， 每组数据有5个整数，分别对应函数f 中(1)、(2)、(3)、(4)、(5)。

输出

输出f的表达式，具体看给出的样例输出，不要有多余的符号。

样例输入

2 3 -3 4 -51 2 3 -4 52 0 2 2 2

样例输出

2+3b-3c+4d-51+2b+3c-4d+52+2c+2d+2

#include<stdio.h> 
#include<string.h>
char a[100000],b[100000],c[100000],d[100000],e[100000];
int main()
{
while(scanf("%s%s%s%s%s",a,b,c,d,e)!=EOF)
{
if(strcmp(a,"1")==0&&strcmp(b,"1")==0&&strcmp(c,"1")==0&&strcmp(d,"1")==0&&strcmp(e,"1")==0)
{
printf("0\n");
continue;
}
if(a[0]=='0');
else
printf("%s",a);
if(b[0]=='0');
else if(b[0]=='-')
{
if(strcmp(b,"-1")==0)
printf("-b");
else
printf("%sb",b);
}
else if(strcmp(b,"1")==0)
printf("+b");
else
printf("+%sb",b);
if(c[0]=='0');
else if(c[0]=='-')
{
if(strcmp(c,"-1")==0)
printf("-c");
else
printf("%sc",c);
}
    else {
    if(strcmp(c,"1")==0)
    printf("+c");
   else
     printf("+%sc",c);
    }
  if(d[0]=='0');
else if(d[0]=='-')
{
if(strcmp(d,"-1")==0)
printf("-d");
else
printf("%sd",d);
}
else if(strcmp(d,"1")==0)
   printf("+d");
   else
  printf("+%sd",d);
  if(e[0]=='0');
else if(e[0]=='-')
printf("%s",e);
else printf("+%s",e);
printf("\n");
}
return 0;
}