poj3370 鸽笼原理

题意：首先输入c n，接着是n个数，问你有没有k个数(1<=k<=n),使得其和能整除c

输出其对应下标，只输出一组符合条件的就行；

分析：http://blog.csdn.net/bigsungod/article/details/41450729

代码：

#include <queue>

#include <stack>

#include <math.h>

#include <vector>

#include <limits.h>

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <functional>

#define N 100010

#define LL long long

#define mem(a) memset(a,0,sizeof(a));

#define mem_1(a) memset(a,-1,sizeof(a));

using namespace std;

int n,c;

int f[N];

int sum[N];

int pos[N];

void work()

{

int j;

memset(pos, -1, sizeof(pos));

pos[0] = 0;

for (int i = 1; i <= n; i++)

if (pos[sum[i]] == -1)

pos[sum[i]] = i;

else

{

for ( j = pos[sum[i]] + 1; j < i; j++)

printf("%d ", j);

printf("%d\n", j);

return;

}

}

int main()

{

while(scanf("%d%d", &c,&n)){

if(c==0&&n==0)

break;

for (int i = 1; i <= n; i++)

scanf("%d", &f[i]);

sum[0] = f[0] = 0;

for (int i = 1; i <= n; i++)

sum[i] = (sum[i - 1] + f[i]) % c;

work();

}

return 0;

}