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 50 300 0
Sample Output
[1,4] [10,10] [4,8] [6,9] [9,11] [30,30]
思路
（（首项+末项）*项数）/2=m以及 末项=首项+项数-1联立方程组可以得到一个关于左区间项，项数，M的一个方程。 假设首项是1，我们代入，很容易得到n(n+1)=2m这个公式（n是项数）。 这里可以把n+1看成n，n^2=2m,n=sqrt(2m); 也就是说项数最多的情况也只有sqrt(2m)。大大减小了枚举长度。另外，(a+a+len)*(len+1)/2 = m => a = m/(len+1)-len/2
#include<stdio.h>
#include<math.h>
int main()
{
int N,M,val,len;
while (~scanf("%d%d",&N,&M) && N && M)
{
len = (int)sqrt(2*M);
while (len--)
{
val = M / (len + 1) - len /2;
if ((2*val+len) * (len+1) / 2 == M)
printf("[%d,%d]\n", val, val+len);
}
printf("\n");
}
return 0;
}