[POJ2356]Find a multiple 题解（鸽巢原理）

##[POJ2356]Find a multiple ###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.

###Solution

1.题目大意即为求n个数中区间和被n整除的第一个区间[l,r]；

2.由鸽巢原理知，n个不同的数会造成n个不同的前缀和，那么至少有两个前缀和关于n是同余的，我们只需找出第一个出现两次的模后前缀和即可；

3.打一个地址标记，记录模值的第一次出现地址，当第二次出现同一模后前缀和时，使l=第一次出现地址，r=第二次出现地址，区间[l,r]即为所求；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std; 
int main(){
	int i,j,l=0,r,n,t[10001]={},a[10001]={},sum[10001]={};
	scanf("%d",&n);
	memset(t,-1,sizeof(t));
	t[0]=0;
	for(i=1;i<=n;++i){
		scanf("%d",&a[i]);
		sum[i]=(sum[i-1]+a[i])%n;
		if(t[sum[i]]!=-1){
			l=t[sum[i]];
			r=i;
			break;
		}
		else t[sum[i]]=i;//zzh大佬的计数方法； 
	}
	printf("%d\n",r-l);
	for(i=l+1;i<=r;++i)printf("%d\n",a[i]);
	return 0; 
}

有关鸽巢原理可以参考我的博客：http://www.cnblogs.com/COLIN-LIGHTNING/p/8439555.html

来源：oschina

链接：https://my.oschina.net/u/4408350/blog/4244857

标签

one number