7E

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]

// 遍历所有子列

#include<stdio.h>
int main()
{
    int n, m, i,j, t, flag;
    while(scanf("%d %d", &n, &m), !(n==0&&m==0))
    {
        for(i=1;i<=n;i++)
        {
            flag=0; t=0;
            for(j=i;j<=n;j++)
            {
                t+=j;
                if(t==m)
                { flag=1; break; }
                if(t>m) break;
            }
            if(flag) printf("[%d,%d]
", i, j);
        }
        printf("
");
    }
    return 0;
}

// 老老实实算吧

#include<stdio.h>
#include<math.h>
int main()
{
    int n, m, i,j;
    while(scanf("%d %d", &n, &m), !(n==0&&m==0))
    {
        for(i=(int)sqrt(2.0*m);i>0;i--)    // a1*n+n*(n-1)/2=(2*a1+(n-1))*n/2=m,
        {                                // 又a1>=1, 则n<sqrt(2*m). 
            j=(2*m/i-i+1)/2;            // a1=(2*m/n-n+1)/2
            if((2*j+i-1)*i/2==m)
                printf("[%d,%d]
", j, i+j-1);
        }
        printf("
");
    }
    return 0;
}