poj3370同样的是抽屉原理

Halloween treats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3862		Accepted: 1470		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

题目大意是给出n和m及m个整数（n<m），要从m中找出任意个数使得其和是n的倍数，输出时就输出每一个数所在的编号

例如对于题目中的数据二，明显可以找出11，2，5，他的和是18，是3的倍数，而11，2，5对应于题目中的数据分别是2，3，4，故结果输出的是2，3，4

当然，除了2，3，4这一个解，还有1，2也是题目的解，但题目要求对于多解的情形只用输出一组解，如果无解，就输出“no sweets”

但根据抽屉原理，可以得出没有无解的情况，

详细证明请看我的关于poj2356的解题报告，这两题长得非常相像

http://www.cnblogs.com/ACShiryu/archive/2011/08/09/poj2356.html

刚开始时没注意到数据范围，TLE和WA分别一次，这题数据明显比2356强，中间有可能超int，故要用__int64，并且cin比scanf慢

参考代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[100000] , mod[100000] ;
int main()
{
    int c , n ;
    while ( scanf("%d%d",&c,&n) , c || n )
    {
        int i , j ;
        for ( i = 0 ; i < n ; i ++ )
            scanf("%d",&a[i]) , mod[i] = -2 ;//将mod初始化为-2
        mod[0]=-1 ;//mod[0]为-1，就是假设存在a[-1]，且a[-1]是n的倍数，这样就可以把两种情况写在一起
        __int64 sum = 0 ;//直接用sum，省去了另开数组的空间
        for ( i = 0 ; i < n ; i ++ )
        {
            sum += a[i] ;
            if ( mod [ sum % c ] != -2 )
            {//如果在i之前有与sum对n同余的数，则可以输出答案，
                for ( j = mod [ sum % c ] + 1 ; j <= i ; j ++ )
                {
                    cout<<j+1;
                    if ( i != j )
                        cout<<' ';
                }
                cout<<endl;
                break;
            }
            mod [sum%c] = i ;//记录余数对应的是i
        }
    }
    return 0;
}