题目链接：
http://poj.org/problem?id=2864
Pascal Library
Description
Pascal University, one of the oldest in the country, needs to renovate its Library Building, because after all these centuries the building started to show the effects of supporting the weight of the enormous amount of books it houses.
To help in the renovation, the Alumni Association of the University decided to organize a series of fund-raising dinners, for which all alumni were invited. These events proved to be a huge success and several were organized during the past year. (One of the reasons for the success of this initiative seems to be the fact that students that went through the Pascal system of education have fond memories of that time and would love to see a renovated Pascal Library.)
The organizers maintained a spreadsheet indicating which alumni participated in each dinner. Now they want your help to determine whether any alumnus or alumna took part in all of the dinners.
Input
The input contains several test cases. The first line of a test case contains two integers N and D indicating respectively the number of alumni and the number of dinners organized (1 <= N <= 100 and 1 <= D <= 500). Alumni are identified by integers from 1 to N. Each of the next D lines describes the attendees of a dinner, and contains N integers Xi indicating if the alumnus/alumna i attended that dinner (Xi = 1) or not (Xi = 0). The end of input is indicated by N = D = 0.
Output
For each test case in the input your program must produce one line of output, containing either the word `yes‘, in case there exists at least one alumnus/alumna that attended all dinners, or the word `no‘ otherwise.
Sample Input
3 3 1 1 1 0 1 1 1 1 1 7 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0
Sample Output
yes no
Hint:
题意：
给你一个n,d。n表示参加聚会的人数，d表示聚会的举办次数。要你求出是否有人出席了全部的聚会，有的话就输出yes，不然输出no。
题解：
简单的模拟，注意一下中间的循环即可。
代码：
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define met(a,b) memset(a,b,sizeof(a))
#define maxn 500+10
int map[maxn][maxn];
int n,m;
int main()
{
while(scanf("%d%d",&n,&m),n!=0||m!=0)
{
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
scanf("%d",&map[i][j]);
int flag=0;
for(int i=0;i<n;i++)
{
int num=0;
for(int j=0;j<m;j++)
{
if(map[j][i]==1)
num++;
}
if(num==m)
{
flag=1;
break;
}
}
if(flag==1)
printf("yes\n");
else
printf("no\n");
}
}