2058 The sum problem 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 contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.
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


#include <cstdio>
#include <cmath>
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]\n", i, i+k-1);
    return 0;