HDOJ 题目2058The sum problem（数学）

最新推荐文章于 2020-04-24 16:49:52 发布

Jogging_Clown

最新推荐文章于 2020-04-24 16:49:52 发布

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本文探讨了一个经典的子序列求和问题，即给定一个由1到N组成的序列，找出所有可能的子序列使得子序列的和等于给定值M。通过数学推导找到最长子序列的长度，并给出了一种有效的解决方案。

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The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15381 Accepted Submission(s): 4620

Problem Description

Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.

Input

Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

Output

For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

Sample Input

Sample Output

  
   [1,4]
[10,10]

[4,8]
[6,9]
[9,11]
[30,30]

Author

8600

Source

校庆杯Warm Up

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思路：

先确定一段和sum＝m的最长的长度，要最长则起点从1开始，根据等差求和，len*(len+1)/2=m;放缩法则len^2<2*m,所以len=pow(2.0*m,0.5);现在根据长度确定起点L，根据等差求和公式，((len+L-1)+L)*len/2=m; => L=(2*m/len+1-len)/2; 判断时反代上式是否成立，是则输出[L,L+len-1。

ac代码

#include<stdio.h>
#include<math.h>
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF,n||m)
	{
		//int len=sqrt((double)2*m);
		int len=(int)pow(2.0*m,0.5);
		if(len>n)
			len=n;
		for(;len>=1;len--)
		{
			int l=(2*m/len-len+1)/2;
			if(((len-1+l)+l)*len/2==m)
				printf("[%d,%d]\n",l,len+l-1);
		}
		printf("\n");
	}
}