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(*m); k>; k--)
         {
             a = m/k-(k-)/;
             if((*a-+k)*k == *m)
                 cout << "[" << a << "," << k+a- << "]" << endl;
         }
     cout << endl;
     }
     return ;
 }