POJ 2356 (抽屉原理)

题目中说：随便输出一组。抽屉原理可以求出取出的数连续的情况。

#include <iostream>
#include<cstdio>
using namespace std;
int a[10005],s[10005];
int n,start,end,ok = 0;
int main()
{
    while(scanf("%d",&n) != EOF )
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
            s[i] = (s[i-1] + a[i])%n;
        }
    for(int  i = 1; i <= n; i++)
    {
        if(s[i] == 0)
        {
            start = 0;
            end = i;
            ok = 1;
            break;
        }
        for(int j = i+1;j <= n;j++)
        {
            if(s[i] == s[j])
            {
                start = i;
                end = j;
                ok = 1;
                break;
            }
        }
        if(ok)
            break;//注意上面的break只是退出了j循环，没有退出i循环
    }
    if(ok)
    {
        printf("%d
",end-start);
        for(int i = start+1;i <= end;i++)
            printf("%d
",a[i]);
    }
    else
        printf("0
");
    }
    return 0;
}