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】
1 #include <iostream> 2 #include <math.h> 3 using namespace std; 4 int main() 5 { 6 __int64 m, n; 7 while(cin >> n >> m, n||m) 8 { 9 __int64 a, k; 10 for(k=(int)sqrt(2*m); k>0; k--) 11 { 12 a = m/k-(k-1)/2; 13 if((2*a-1+k)*k == 2*m) 14 cout << "[" << a << "," << k+a-1 << "]" << endl; 15 } 16 cout << endl; 17 } 18 return 0; 19 }