hdu2058

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29632    Accepted Submission(s): 8887

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]

 

分析：设满足条件的区间为[a,a+k-1],那么M=a+(a+1)+......+(a+k-1)=k*a+k*(k-1)/2,

然后枚举k,算出a,记录结果。

 

#include<cstdio>
long long a[1000000],num[1000000];
int main()
{
    long long N,M;
    while(scanf("%lld%lld",&N,&M))
    {
        if(!N&&!M) break;
        int cnt=0;
        long long k=1,temp;
        while(true)
        {
            temp=M-k*(k-1)/2;
            if(temp<=0) break;
            if(temp%k==0&&temp/k+k-1<=N)
            {num[cnt]=k-1;a[cnt++]=temp/k;}
            k++;
        }
        for(int i=cnt-1;i>=0;i--)
        printf("[%lld,%lld]
",a[i],a[i]+num[i]);
        printf("
");
    }
    return 0;
}