【POJ 2356】 Find a muliple 鸽笼原理-优快云博客

本文链接：https://blog.youkuaiyun.com/NLt_XXy/article/details/79887943

本文介绍了一个关于从一组正整数中找到若干个数，使得这些数的总和能够被该组数的数量整除的问题。通过计算前缀和并利用鸽笼原理，找到了有效的解决方案。

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Find a multiple

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8716 Accepted: 3765 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

题解

要找出整除 $n$ 的数，一共有 $n$ 个数，我们计算他们的前缀和，
一共有 $n$ 个前缀和，这 $n$ 个数 $\mod n$ 的余数，一共有 $n$ 种情况:
如果余数为 $0$ ,我们直接从1开始输出到这个数就可以了
如果余数不为，则必有两个数的余数相同(鸽笼原理)，我们减一下就得到了结果

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
const int MAXN=1e4+10;
using namespace std;
int a[MAXN],MOD[MAXN],sum,n;
bool ok;
void print(int l,int r)
{
    printf("%d\n",r-l+1);
    for(int i=l;i<=r;i++)
        printf("%d\n",a[i]);
    return ;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
    {
        sum+=a[i];
        if(sum%n==0&&(!ok)) 
        {
            print(1,i);
            ok=1;
        }
        if(MOD[sum%n]!=0&&(!ok))
        {
            print(MOD[sum%n]+1,i);
            ok=1;
        }
        if(MOD[sum%n]==0)
            MOD[sum%n]=i;
    }
    return 0;
}