求序列的和，杭电0j-2058

原题地址：http://acm.hdu.edu.cn/showproblem.php?pid=2058

【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]

【AC代码】

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    __int64 m, n;
    while(cin >> n >> m, n||m)
    {
        __int64 a, k;
        for(k=(int)sqrt(2*m); k>0; k--)
        {
            a = m/k-(k-1)/2;
            if((2*a-1+k)*k == 2*m)
                cout << "[" << a << "," << k+a-1 << "]" << endl;
        }
    cout << endl;
    }
    return 0;
}