Poj2356find a multiple Pigeon nest Principle

There must be a continuous K number, so that their abilities and abilities are divisible by N. Set a [I] As the prefix and

A [1] % N, a [2] % N ,..., the range of a [n] % N value is <n, so there are n numbers smaller than N. Two identical numbers will certainly appear, indicating, A part of the second number that is more than the first number must be divisible by N.

Pay attention to the handling of prefix and occurrence of 0 in the case.

#include<iostream>#include<cstdio>#include<cstring>#include<map>#include<vector>#include<stdlib.h>using namespace std;typedef long long LL;int main(){    int n;    int a[22222];    int b[22222];    int vis[22222];    while (cin >> n){        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        memset(vis, 0, sizeof(vis));        for (int i = 1; i <= n; i++)            cin >> a[i], b[i] = a[i];        for (int i = 1; i <= n; i++){            b[i] += b[i - 1]; b[i] %= n;        }        int gg = 0;        for (int i = 1; i <= n;i++)        if (b[i] == 0){            cout << i << endl;            for (int j = 1; j <= i; j++)                cout << a[j] << endl;            gg = 1; break;        }        if (gg) continue;        int flag = 0;        for (int i = 1; i <= n; i++){            if (flag) continue;            if (!vis[b[i]]){                vis[b[i]] = i;            }            else{                cout << i - vis[b[i]] << endl;                for (int j = vis[b[i]]+1; j <= i; j++)                    cout << a[j] << endl;                break;            }        }    }    return 0;}

 

Poj2356find a multiple Pigeon nest Principle