HDOJ 2058 The sum problem(数学问题）

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8287    Accepted Submission(s): 2530

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

20 10
50 30
0 0

 

Sample Output

[1,4]
[10,10]

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

 

我本来的想法是考虑子列的起点和终点，分别以s和e表示，由等差数列求和公式有（s+e）*（e-s+1）/2==M（1式），化为e*(e+1)-s*(s-1)==2*M, so , e=(int)sqrt(2*M+s*(s-1)),将得到的e再代回1式，成立则[s,e]满足条件。

但是，2*M+s*(s-1)太大……

后来参考网上的一个算法，不考虑子列的终点，而是考虑子列的起点和子列元素的个数，分别记为i，j。由等差数列求和公式，得(i+(i+j-1))*j/2==M ，即（2*i+j-1）*j/2==M（2式），故得i=（2*M/j-j+1）/2，将i，j代回2式，成立则[i,i+j-1]满足条件。注意j最小为1，而由2式，得（j+2*i）*j=2*M，而i>=1，故j*j<=(int)sqrt(2*M).

AC code:

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	int N,M,i,j;
	while(scanf("%d%d",&N,&M) && M+N)
	{ 
	 	for(j=pow(2.0*M,0.5);j>0;j--)
		{
			i=(2*M/j-j+1)/2;
			if(j*(j+2*i-1)/2==M) printf("[%d,%d]\n",i,i+j-1);
		}
		printf("\n");					  
 	} 	
	return 0;	
}