hdu2058 The sum problem(枚举~~等差数列求和公式)

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17697    Accepted Submission(s): 5275

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<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define maxn 1000000
__int64 N,M;
int main()
{
    while(scanf("%I64d%I64d",&N,&M)!=EOF&&N&&M)
    {
        for(__int64 k=(int)sqrt(2*M);k>=1;k--)
        {
            __int64 a1=M/k-(k-1)/2;
            if((2*a1+k-1)*k==2*M)
            {
                printf("[%I64d,%I64d]\n",a1,a1+k-1);
            }
        }
        printf("\n");
    }
    return 0;
}
//根据等差数列求和，m=(2*a1+k-1)*k/2,k表示数列的项数，a1表示首项。
//枚举k(1<=k<=sqtrt(m))