HDU 2356 Find a multiple(鸽巢原理）

Find a multiple

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8367 Accepted: 3636 Special Judge

Description
The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This numbers are not necessarily different (so it may happen that two or more of them will be equal). Your task is to choose a few of given numbers ( 1 <= few <= N ) so that the sum of chosen numbers is multiple for N (i.e. N * k = (sum of chosen numbers) for some natural number k).

Input
The first line of the input contains the single number N. Each of next N lines contains one number from the given set.

Output
In case your program decides that the target set of numbers can not be found it should print to the output the single number 0. Otherwise it should print the number of the chosen numbers in the first line followed by the chosen numbers themselves (on a separate line each) in arbitrary order.

If there are more than one set of numbers with required properties you should print to the output only one (preferably your favorite) of them.

Sample Input

5
1
2
3
4
1

Sample Output

2
2
3

Source
Ural Collegiate Programming Contest 1999

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题意：给出一个数n，输入n个数，找出连续排列的一串数字，满足为n的倍数，输出数字串长度和数字串，每个数字一行。

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鸽巢原理的入门题，水，但是错了很多次，看了模版，细节问题，找了很久bug。。。
暴力还容易一些~

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鸽巢简单理解就是：假如有n+1只鸟，有n个巢，这些鸟都回巢，那么肯定有一个巢的鸟多于两只。

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#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 11111int n, i, j;int a[N] = {0}, b[N] = {0}, vis[N];void run(){    memset( vis, -1, sizeof( vis ));    vis[0] = 0;    for( i = 1 ; i <= n ; i ++ ){///        if( vis[b[i]] == -1 )            vis[b[i]] = i;        else{            cout<<i-vis[b[i]]<<endl;///            for( j = vis[b[i]]+1; j <= i ; j ++ ){                cout<<a[j]<<endl;            }            return;        }    }}int main(){    //freopen( "in.txt", "r", stdin );    scanf( "%d", &n );    for( i = 1 ; i <= n ; i ++ ){///从1开始输入，上面容易输出        scanf( "%d", &a[i] );    }    //b[0] = a[0] % n;///    b[0] = a[0] = 0;    for( i = 1 ; i <= n ; i ++ ){        b[i] = (a[i] + b[i-1]) % n;    }    run();    return 0;}