杭电2058 The sum problem

最新推荐文章于 2021-01-29 14:27:20 发布

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最新推荐文章于 2021-01-29 14:27:20 发布

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The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19582 Accepted Submission(s): 5772

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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我想这就叫算法吧，利用高斯求和，进行反解，由于解出a1比较难，就解出了n：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;//2*s=(2*a1+n-1)*n;
long i,j,k,l,m,n;
double help;
int main()
{
	while(scanf("%ld%ld",&m,&n),m+n)
	{
		help=sqrt(2*n+1);
		if(m<help)
		help=m;
		for(i=help+1;i>=1;i--)
		{
			l=(n*2+i-i*i)/2/i;
			if(i*i-i+2*l*i==2*n&&l>0&&(i%2==0||(2*l+i-1)%2==0))
			printf("[%ld,%ld]\n",l,i+l-1);
		}
		printf("\n");
	}
}