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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.

InputInput contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0. 
OutputFor 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]

数学思维题目，，看了一下大佬的博文。。感觉 还复杂啊！！

#include<iostream>
#include<algorithm>
#include<string> 
#include<cstdio>
#include<math.h>
using namespace std;
/*
思路:
区间项数为i
起始位置 a,终点为b则b=a+i-1;
这一段区间的和为sum=(a+b)*i/2=(a+a+i-1)*i/2=m;
a最小为1，则（1+1+i-1）*i/2<=m即(i+1)*i<=2m , i有一定小于sqrt(2m)
即 0<i<sqrt(2m);判断条件就是 (a+a+i-1)*i==2*m;
*/
int main()
{
    int n,m,a,b;
    while(cin>>n>>m)
    {
        if(n==0 && m==0)
            break;
        for(int i=sqrt(2*m);i>=1;i--)//最小为1项，就是[m,m] 
        {
            a=m/i-(i-1)/2;//等差公式 即 m=a*i+i*(i-1)/2-----m/i=a+(i-1)/2---a=m/i-(i-1)/2;
            if((a+a+i-1)*i==2*m)
            {
                printf("[%d,%d]
",a,a+i-1);
            }
        }
        printf("
");
    }
    return 0;
}