[POJ3370]&[HDU1808]Halloween treats 题解（鸽巢原理）

[POJ3370]&[HDU1808]Halloween treats

Description

-Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too late. To avoid conflicts, the children have decided they will put all sweets together and then divide them evenly among themselves. From last year's experience of Halloween they know how many sweets they get from each neighbour. Since they care more about justice than about the number of sweets they get, they want to select a subset of the neighbours to visit, so that in sharing every child receives the same number of sweets. They will not be satisfied if they have any sweets left which cannot be divided.
Your job is to help the children and present a solution.

-Input：The input contains several test cases.
The first line of each test case contains two integers c and n (1 ≤ c ≤ n ≤ 100000), the number of children and the number of neighbours, respectively. The next line contains n space separated integers a1 , ... , an (1 ≤ ai ≤ 100000 ), where ai represents the number of sweets the children get if they visit neighbour i.
The last test case is followed by two zeros.

-Output：
For each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of ai sweets). If there is no solution where each child gets at least one sweet print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them.

Solution

1.本题与Find a multiple（题解随笔：http://www.cnblogs.com/COLIN-LIGHTNING/p/8481478.html）基本相同，只需多组数据处理，数据一定要先全部读完再处理；

2.由于n<m，那么至少有两个邻居给的糖果数前缀和关于孩子数同余，取模后相同的前缀和后出现的地址减去前面的地址即可；

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，注意题目要求输出的是可能的一组地址；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std; 
int i,j,l,r,t[100001]={},a[100001]={},sum[100001]={};
void cul(int n,int m){                     //分组处理数据；
	l=0;
	memset(t,-1,sizeof(t));
	memset(a,0,sizeof(a));
	memset(sum,0,sizeof(sum));
	t[0]=0;
	for(i=1;i<=m;++i) scanf("%d",&a[i]);
	for(i=1;i<=m;++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;          //计录模值首次出现的地址； 
	}
	for(i=l+1;i<=r;++i)printf("%d ",i);  //注意输出地址；
	printf("\n");
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	while(n!=0){
		cul(n,m);
		scanf("%d%d",&n,&m);
	}
	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