HDU_2058——等差数列,子集,集合长度判断

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]

 /*
 设j为子串的长度，i为子串的第一个数
 m=[j*(i+i+(j-1)*1)]/2
 2*m/j-j=i+i-1----2*i=2*m/j-j+1
 因为i+i-1>0
 所以2*m>j*j----j<sqrt(2m)
 */
 #include <cstdio>
 #include <cmath>
 int main()
 {
    int n,m,i,j;
    while(~scanf("%d%d",&n,&m),(n||m))
       {
          for(j=(int)sqrt(*(double)m);j>=;j--)//枚举长度
             {
                i=(*m/j-j+)/;//由长度获得的子串第一个元素值
                if((j*(i+i+j-))/ == m)//等差数列前n项的和，还有通项公式
                   printf("[%d,%d]\n",i,i+j-);
             }
          printf("\n");
       }
    return ;
 }