The sum problem

The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 18248 Accepted Submission(s): 5421



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]

Author

8600



Source

校庆杯Warm Up


这是一个数学问题，直接求会超时！

利用数学公式中的等差公式就可以AC了！

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
/*
M=Sn =a1*n+(n-1)n/2=a1*N+(N-1)N/2,
可得a1*N=M-(N-1)N/2;
*/
int main()
{
    int n,m,sum,i;
    while(cin>>n>>m&&(n+m))
    {
       // for(i=n;i>=1;i--)
       for(i=(int)sqrt(2*m);i>=1;i--)
        {
            sum=m-i*(i-1)/2;
            if(sum%i==0)
            {
                cout<<"["<<sum/i<<","<<sum/i+i-1<<"]"<<endl;
            }
        }
        cout<<endl;
    }
}