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

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

 

Sample Output

Case #1: Yes
Case #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]

题目描写叙述

声明：这道题没有涉及不论什么算法！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，分别相应函数f 中(1)、(2)、(3)、(4)、(5)。

输出

输出f的表达式，详细看给出的例子输出，不要有多余的符号。

例子输入

2 3 -3 4 -5
1 2 3 -4 5
2 0 2 2 2

例子输出

2+3b-3c+4d-5
1+2b+3c-4d+5
2+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;
}