3370 Halloween treats 抽屉原理给定n个数，求其中的任意一个子集满足集合中的每个元素值加和正好是c的倍数

最新推荐文章于 2020-12-24 13:56:08 发布

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本文介绍了一种解决万圣节孩子们如何公平分配从邻居那里收集到的糖果的算法。该算法通过挑选合适的邻居访问来确保每个孩子都能获得相同数量的糖果，并提供了具体的实现代码。

Halloween treats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3242		Accepted: 1205		Special Judge

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 , ... , a_n (1 ≤ a_i ≤ 100000 ), where a_i 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 a_i 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.

Sample Input

4 5
1 2 3 7 5
3 6
7 11 2 5 13 17
0 0

Sample Output

3 5
2 3 4

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[110010],vis[110010],s[110010];
int main()
{
    int c,n;
    while(scanf("%d%d",&c,&n)==2&&c)
    {
        int pos,len=0;
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            s[i]=(s[i-1]+a[i])%c;
            if(len) continue;//已经找到解
            int mod=s[i];
            if(mod==0)
            {
                pos=1,len=i;
                continue;
            }
            if(vis[mod]==0)
            {
                vis[mod]=i;
            }
            else
            {
                len=i-vis[mod];
                pos=vis[mod]+1;
            }
        }
        printf("%d",pos);
        for(int i=pos+1;i<pos+len;i++) printf(" %d",i);printf("/n");
    }
    return 0;
}