武汉科技大学计算机学院菜鸟杯：The sum problem （杭电2058)

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10444    Accepted Submission(s): 3180

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]

 

 

代码如下：

/*
找一个子序列的和便是M，那么这个子序列可以算作a+1， a+2， ... ， a+d
这时，d就为这个序列的长度，肇端数字就是a+1，而这个序列的和即M=a*d+(1+d)*d/2;
得出d*d<2 * m，从而可以列举d，策画出a
*/

#include<stdio.h>
#include <math.h>

int main()
{
    int i,j,m,n;
    while(scanf("%d%d",&n,&m),n&&m)
    {
        for(i=sqrt(2*m);i>0;i--)
        {
            j=m-(i+i*i)/2;
            if(j%i== 0)
            printf("[%d,%d]\n",j/i+1,j/i+i);
        }
        printf("\n");
    }
    return 0;
}