The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22089    Accepted Submission(s): 6508


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]
Author 8600
Source

高斯公式的應用，如果暴力會超時。

題解：

程式碼如下：

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
	long long n,m;
	long long a,b;
	while (~scanf("%lld %lld",&n,&m) && (n || m))
	{
		for (int i = sqrt((double)(2*n)) ; i > 0 ; i--)
		{
			if ((2 * m) % i == 0 && ((2 * m / i) + i - 1) % 2 == 0)
			{
				a = ((2 * m / i) + i - 1) / 2;
				b = 2 * m / i - a;
				if (a > b)
					swap(a,b);
				if (a > 0)
					printf ("[%d,%d]\n",a,b);
			}
		}
		printf ("\n");
	}
	return 0;
}