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(2*(double)m);j>=1;j--)//枚举长度 
            {
               i=(2*m/j-j+1)/2;//由长度获得的子串第一个元素值
               if((j*(i+i+j-1))/2 == m)//等差数列前n项的和，还有通项公式
                  printf("[%d,%d]
",i,i+j-1);
            }
         printf("
");
      }
   return 0;   
}