1.6 [hduoj] 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]

#include<stdio.h>
#include<math.h>
//高斯公式 和=（首项+末项）*项数/2
void main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF&&!(n==0&&m==0))
    {
        for(int i=(int)(sqrt(2*m));i>=1;i--)
        {
            int a=m/i-(i-1)/2;
            int b=i+a-1;
            if((a+b)*i==2*m)
                printf("[%d,%d]\n",a,b);
        }
        printf("\n");
    }
}

第一遍直接用递推加高斯公式，没有任何技巧，额。时间就超了。时间复杂度大概有n²

第二遍，用了数学方法，求出项数的限制条件，根据项数来进行递推。就只有一重循环了。