POJ1032学习-数论

最新推荐文章于 2019-02-18 16:12:00 发布

原创最新推荐文章于 2019-02-18 16:12:00 发布 · 212 阅读

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本文针对一个关于议会工作天数最大化的算法问题进行了探讨，并提供了一种有效的解决方案。通过合理的分组，使得每一天都能有不同的代表团成员构成调解委员会，从而确保议会能够持续运作尽可能长的时间。

Parliament

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19708		Accepted: 8334

Description

New convocation of The Fool Land's Parliament consists of N delegates. According to the present regulation delegates should be divided into disjoint groups of different sizes and every day each group has to send one delegate to the conciliatory committee. The composition of the conciliatory committee should be different each day. The Parliament works only while this can be accomplished.
You are to write a program that will determine how many delegates should contain each group in order for Parliament to work as long as possible.

Input

The input file contains a single integer N (5<=N<=1000 ).

Output

Write to the output file the sizes of groups that allow the Parliament to work for the maximal possible time. These sizes should be printed on a single line in ascending order and should be separated by spaces.

Sample Input

Sample Output

3 4

借鉴了前辈的思路：https://blog.youkuaiyun.com/aidway/article/details/50896811

修改了一下最后的循环方法。

代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int s[520];
int main()
{
	int N;
	while(scanf("%d",&N)!=EOF)
	{
		memset(s,0,sizeof(s));
		int sum=2;
		int i;
		s[1]=2;
		for(i=2;i<=N;i++)
		{
			if(sum+s[i-1]+1<=N)
		 	{
		 		s[i]=s[i-1]+1;
		 		sum+=s[i];
		 	}
		 	else
		 	{
		 		sum=N-sum;
		 		i--;
		 		break;
		 	}
		}
		while(sum>0)
		{
			for(int j=i;j>0;j--)
			{
				if(sum>0)
				{
					s[j]++;
					sum--;
				}
				else
					break;
			}
		}
		for(int j=1;j<=i;j++)
		{
			printf("%d",s[j]);
			if(j<i)
				printf(" ");
			else
				printf("\n");
		}
	}
	return 0;
}