HDU 2058 The sum problem

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20505    Accepted Submission(s): 6023

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]

 

 

 

解析：直接枚举会超时，因此枚举时要缩小范围。根据等差求和公式Sn = n*a₁ +n*(n-1)/2*d,

此处Sn = m,d = 1,则m = n*a₁+n*(n-1)/2。整理得2*m = n*n+(2*a₁-1)*n。可得：

n<sqrt(2.0*m)。

 

 

 

#include <cstdio>
#include <cmath>

int n,m;

int main()
{
    while(scanf("%d%d",&n,&m), n){
        for(int num = (int)sqrt(m*2.0); num>0; --num){
            int numa1 = m-num*(num-1)/2;
            if(numa1%num == 0)
                printf("[%d,%d]
",numa1/num,numa1/num+num-1);
        }
        printf("
");
    }
    return 0;
}