POJ2270&&Hdu1808 Halloween treats 2017-06-29 14:29 40人阅读 评论(0) 收藏

Halloween treats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8565		Accepted: 3111		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

Source

Ulm Local 2007

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题意：给你n个数和c，取出一些数，使得这些数字和对c取模为0，输出这些数的位置

解题思路：关键点在c<=n，若前缀和对c取模为0，则直接输出所有数，否则的话，因为c<=n，所以一定会出现同余的情况，同余的两个位置之间所有的数就是答案

注意：POJ可能会卡常数

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[100005];
int cnt[100005];
int main()
{
    int m,n;
    while(~scanf("%d%d",&n,&m)&&(m||n))
    {

        for(int i=1; i<=m; i++)
        {
            scanf("%d",&a[i]);
        }
        int sum=0;
        memset(cnt,-1,sizeof cnt);
        for(int i=1; i<=m; i++)
        {
            sum+=a[i];
            if(sum%n==0)
            {
                for(int j=1; j<i; j++)
                {
                    printf("%d ",j);
                }
                printf("%d\n",i);
                break;
            }
            else
            {
                sum%=n;
                if(cnt[sum]>0)
                {
                    for(int j=cnt[sum]+1; j<i; j++)
                    {
                        printf("%d ",j);
                    }
                    printf("%d\n",i);
                    break;
                }
                else
                    cnt[sum]=i;
            }
        }
    }
    return 0;
}