POJ-3370 Halloween treats （抽屉原理）

本是古典何须时尚 2022-06-12 05:43 128阅读 0赞

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 ***a*1 , ... , *a**n*** (*1 ≤ ai ≤ 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

这道题的题意就是输入两个数c,n,然后下一排输入n个数，要在n个数中任意挑几个数出来，使其总和为c的倍数。

这道题我们可以利用抽屉原理，即10个苹果放入9个抽屉，怎么放都会有一个抽屉里有2个苹果，我们取s\[i\]为前i个数的和模c后的值，那么这题的n是大于c的，我们取s\[i\]为前i个数的和模c后的值,总有两个数s\[p\],s\[q\]模c后的值相等，那么p到q就是我们所求的答案。

AC代码如下

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <cmath>
    #include <vector>
    #include <queue>
    #include <map>
    #include <algorithm>
    #include <set>
    #include <functional>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    const int INF = 1e9 + 5;
    const int MAXN = 1e6 + 5;
    const int MOD = 1e8 + 7;
    const double eps = 1e-7;
    const int MAXNUM = 1000;
    
    int p[100005];
    int s[100005];
    int place[100005];
    int main()
    {
    	int m, n;
    	int flag;
    	int ans;
    	int anss;
    	while (scanf("%d%d",&m,&n)!=EOF)
    	{
    		flag = 0;
    		memset(place, 0, sizeof(place));
    		memset(s, 0, sizeof(s));
    		if (m == 0 & n == 0)
    			break;
    		for (int i = 1; i <= n; i++)
    		{
    			scanf("%d", &p[i]);
    			if (flag != 0)
    				continue;
    			p[i] = p[i] % m;
    			if (p[i] == 0)//如果当前这个数已经是c的倍数了
    			{
    				ans = i;
    				flag = 1;
    			}
    			s[i] = (s[i - 1] + p[i]) % m;
    			if (s[i] == 0)//如过当前的s[i]已经是c的倍数了
    			{
    				ans = i;
    				flag = 2;
    			}
    			if (place[s[i]]!=0)//利用抽屉原理找到了结果
    			{
    				anss = place[s[i]];
    				ans = i;
    				flag = 3;
    			}
    			place[s[i]] = i;
    		}
    		if (flag == 1)
    		{
    			printf("%d\n", ans);
    		}
    		if (flag == 2)
    		{
    			printf("1");
    			for (int i = 2; i <= ans; i++)
    			{
    				printf(" %d", i);
    			}
    			printf("\n");
    		}
    		if (flag == 3)
    		{
    			printf("%d", anss+1);
    			for (int i = anss+2; i <= ans; i++)
    			{
    				printf(" %d", i);
    			}
    			printf("\n");
    		}
    		if (flag == 0)
    			printf("no sweets\n");
    	}
    }