2058 The sum problem

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]

 

 参考链接：The sum problem

最终AC代码：

#include <cstdio>
#include <cmath>
//难点一：想到用等差数据求和的思想；难点二：从长度求首项，而不是从首项求长度思想 
int main(){
    int i, k, n, m;
    while(scanf("%d %d", &n, &m) != EOF){
        if(n==0 && m==0) break;
        for(k=sqrt(2*m); k>0; k--){ //k是区间的长度 
            i = (2*m - k*(k-1)) / (2*k); //i表示首项 
            if(k*i+k*(k-1)/2 == m){ //找到满足条件的区间 
                printf("[%d,%d]
", i, i+k-1);
            }
        }
        printf("
");
    }
    return 0;
}