The sum problem

The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7877 Accepted Submission(s): 2422

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]

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

 

 

这题早就推了公式，只是可能有些地方没有简化，一直出不来，然后今天看了白神的代码，知道了解题思路，首先，把i看作0， 则可求出i的最大值，然后暴力i。