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

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.

（（首项+末项）*项数）/2=m以及 末项=首项+项数-1联立方程组可以得到一个关于左区间项，项数，M的一个方程。假设首项是1，我们代入，很容易得到n(n+1)=2m这个公式（n是项数）。 这里可以把n+1看成n，n^2=2m,n=sqrt(2m); 也就是说项数最多的情况也只有sqrt(2m)。大大减小了枚举长度。另外，(a+a+len)*(len+1)/2 = m => a = m/(len+1)-len/2

#include<stdio.h>#include<math.h>int main(){    int N,M,val,len;    while (~scanf("%d%d",&N,&M) && N && M)    {		len = (int)sqrt(2*M);        while (len--)        {            val = M / (len + 1) - len /2;            if ((2*val+len) * (len+1) / 2 == M)                printf("[%d,%d]\n", val, val+len);        }        printf("\n");    }    return 0;}