POJ 2356 Find a multiple (在n个数组成的数列中找出若干个连续数使它们之和能被n整除）

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本文介绍了一种算法，用于从一组正整数中找出若干个数，使得这些数的总和能够被这组数的数量整除。算法通过计算累积和并对N取模来寻找符合条件的子集。

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

Sample Output

2
2
3

题意：从n个数找出一些数，他们的和能被n整除

解：能被整除即sum%n==0

那么设ans[i]为从1到i的和那么总共有n个ans，他们都mod n

第一种情况有一个ans[i]%n==0 那么答案就是从 1到i

第二种情况没有ans[i]%n==0的那么肯定会有两个的余是相同的 ans[i]%n == ans[j]%n （因为余值从0到n-1 ，而有n个余值），答案就是从 i+1 到 j

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<stack>
#include<vector>
#include<queue>
#include<map>
#define nn 500
#define mm 100001
#define inff 0x3fffffff
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
int n;
long long sum;
int a[10011];
int v[10011];
int st,ed,oo;
int main()
{
   while(~scanf("%d",&n))
   {
       sum=0;
       oo=0;
       memset(v,0,sizeof(v));
       for(int i=1;i<=n;i++)
       {
           scanf("%d",&a[i]);
           if(oo)
            continue;
           sum+=a[i];
           if(sum%n==0)
           {
               st=1;
               ed=i;
               oo=1;
             continue;
           }
           if(!v[sum%n])
               v[sum%n]=i;
           else
           {
               st=v[sum%n]+1;
               ed=i;
               oo=1;
           }
       }
       printf("%d\n",ed-st+1);
       for(int i=st;i<=ed;i++)
        printf("%d\n",a[i]);
   }
}