The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 43443    Accepted Submission(s): 13238

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]

代码：

#include<stdio.h>
#include<math.h>
int n,m;
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        if( n == 0 && m == 0)
            break;
        for(int i = sqrt(2 * m); i >= 1; i --)
        {
            int a = (m - ((i - 1) * i) / 2) / i;
            if(m == a * i + (i * (i - 1)) / 2)
                printf("[%d,%d]\n",a,a+i-1);
        }
        printf("\n");
    }
    return 0;
}