抽屉原理-poj3370-Halloween treats

最新推荐文章于 2020-04-20 17:40:35 发布

jadeyansir

最新推荐文章于 2020-04-20 17:40:35 发布

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本文链接：https://blog.youkuaiyun.com/jadeyansir/article/details/76406993

本文探讨了在万圣节期间如何确保孩子们公平地分配从邻居那里收集到的糖果的问题。通过算法解决如何选择访问哪些邻居以确保每个孩子都能得到相同数量的糖果，并且不会有剩余的糖果。

Halloween treats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8755		Accepted: 3172		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

题意：给出c和n，接下来n个数，求任意的几个数的和为c的倍数，输出任意一组答案；

抽屉原理 n个抽屉放大于n个苹果至少有一个抽屉有大于等于2个苹果

地址点击打开链接

会长代码

#include<cstdio>
#include<vector>
#include<algorithm>
#define MAX 100000
using namespace std;
int sum[MAX+5];
vector<int> d[MAX+5];
int main()
{
	int mod,n;
	while (~scanf ("%d%d",&mod,&n) && (mod!=0 && n!=0))
	{
		for (int i = 0 ; i < mod ; i++)
		{
			d[i].clear();
		}
		for (int i = 1 ; i <= n ; i++)
		{
			int t;
			scanf ("%d",&t);
			sum[i] = sum[i-1] + t;
			sum[i] %= mod;
			d[sum[i]].push_back(i);
		}
		int st,endd;
		if (d[0].size() > 0)
		{
			st = 1;
			endd = d[0][0];
		}
		else
		{
			for (int i = 1 ; i < mod ; i++)
			{
				if (d[i].size() >= 2)
				{
					st = d[i][0]+1;
					endd = d[i][1];
					break;
				}
			}
		}
		while (st <= endd)
		{
			printf ("%d%c",st,st == endd ? '\n' : ' ');
			st++;
		}
	}
	return 0;
}

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int s[100050];
struct P
{
	int mod , num;
}s[100050];
bool cmp(P a,P b)
{
	if(a.mod ==b.mod )
	return a.num <b.num ;
	return a.mod <b.mod ;
}
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i,j,k=-1,l;
		long long sum =0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&s[i]);
			sum+=s[i];//对前i个数累加； 
			s[i].mod =sum%c;
			s[i].num =i+1;
			if(k==-1&&s[i].mod ==0)//前i个数的和是c的倍数 
			k=i;
		}
		if(k==-1)//没有找到这样的数列 
		{
			sort(s,s+n,cmp);//余数排序 
			for(i=0;i<n-1;i++)
			{
				if(k==-1&&s[i].mod ==s[i-1].mod )// 如果两个余数相等，则说明他们原来和是c的倍数 
				{
					k=s[i].num ;
					l=s[i+1].num ;
				}
				if(k==-1)
				printf("\n");
				else
				{
					for(i=k+1;i<l;i++)
					printf("%d",i);
					printf("%d\n",l);
				}
			}
		}
		else
		{
			for(i=0;i<k;i++)
			printf("%d",i);
			printf("%d\n",k+1);
		}
	}
	return 0;
}