The sum problem

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]分析：这是杭电2058题，一开始我用两重遍历，反复tle超时。没办法，去网上找了优化方法。i代表首项，j代表项数。b = (a首 + a末) *项数/2 = (i + i + j - 1)*j/2所以（i + i + j - 1） * j = 2 * b. 所以 j < sqrt(b*2)

同时利用等差数列的求和公式判定一次i 和 j 是否满足，这样能把时间复杂度降低成O(n),果断ac。

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<string.h>
using namespace std;
int main()
{
    double a, b;
    while (cin >> a >> b, a+b)
    {
        int j, i;
        for (j =(int)sqrt(b*2); j > 0; j--)
        {
            i =(int)((2*b)/j + 1 -j)/2;
            if ((i + i + j - 1)*j/2 == b)
                cout << "[" << i << "," << i+j-1 << "]" << endl;
        }
        printf("\n");
    }
    return 0;
}