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]
代码：
#include <bits/stdc++.h>
using namespace std;
int n,m;
int main() {
while(~scanf("%d%d", &n, &m)) {
if( n == 0 && m == 0) break;
for(int i = sqrt(2 * m); i >= 1; i --) {
int a = (m - ((i - 1) * i) / 2) / i;
if(m == a * i + (i * (i - 1)) / 2)
cout<< '[' << a << ',' << a + i - 1 << ']' <<endl;
}
printf("\n");
}
return 0;
}