poj2356

鸽巢原理的ppt：

http://wenku.baidu.com/view/a2f20a2458fb770bf78a55b9.html

别人总结的鸽巢原理：

http://blog.csdn.net/acm_cxlove/article/details/7432166

鸽巢原理：用s1,s2,s3....sk...sn表示从第一个数到第k个数（包括k）的和如果sk能整除n，就直接给出结果，否则，将s1到sn对n的余数求出必有两个余数相同，即si%n == sj%n； 也就(sj-si)%n==0;所以 i+1 到 j的数的和 就是所求；#include <iostream>#include <cstdio>using namespace std;const int maxn = 10000+10;int a[maxn];int sum[maxn];int b[maxn];int main(){    int n;    while(cin >> n)    {        sum[0] = 0;        int flag = 0;        for(int i = 1; i <= n; i++) scanf("%d", a+i);        for(int i = 1; i <= n; i++)        {            //scanf("%d", a+i);            if(a[i]%n==0)            {                printf("1\n%d\n", a[i]);                flag = 1;                break;            }            sum[i] = sum[i-1] + a[i];            if(sum[i]%n==0)            {                printf("%d\n", i);                for(int j = 1; j <= i; j++)                    printf("%d\n", a[j]);                flag = 1;                break;            }            b[i] = sum[i]%n;        }        if(flag) continue;        for(int i = 1; i <= n; i++)            for(int j = 1; j < i; j++)            {                if(b[i] == b[j])                {                    printf("%d\n", i-j);                    for(int k = j+1; k <= i; k++)                        printf("%d\n", a[k]);                }            }    }}

poj3370

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 100000+10;__int64 a[maxn] = {0}, b[maxn];int main(){    int n, m;    while(scanf("%d%d", &n, &m)!=EOF&&n+m)    {        memset(b, 0, sizeof(b));        bool flag = false;        for(int i = 1; i <= m; i++)        {            int tot;            scanf("%d", &tot);            if(flag) continue;            a[i] = a[i-1] + tot;            if(a[i]%n==0)            {                for(int j = 1; j < i; j++)                    printf("%d ", j);                printf("%d\n", i);                flag = true;            }            else if(b[a[i]%n])            {                for(int j = b[a[i]%n]+1; j < i; j++)                    printf("%d ", j);                printf("%d\n", i);                flag = true;            }            else b[a[i]%n] = i;        }    }}