***HDoj 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]

 

 

Author

8600

 

 

Source

校庆杯Warm Up

 

 

Recommend

linle   |   We have carefully selected several similar problems for you:  2059 2062 2060 2072 2061 

 

 

 

 

 

如果按照题意暴力，会超时

根据前n项和公式 Sn = na₁   + n(n-1)/2;

na₁ = Sn - n(n-1)/2 ;

此时假设已知前n项和为m，则

na₁ = m - n(n-1)/2 ;

a₁ =( m - n(n-1)/2  ) / n ;

如果 m - n(n-1)/2 可以被n整除，那么初项 a₁就可以求得出来，子序列末项是 a₁ + n - 1;

 

要注意，如果n从输入的N开始遍历到1仍然会超时

有题目可知： m - n(n-1)/2   >0

可推出  n <√2m

从 √2m 开始遍历，这样就不会超时

 

C语言代码如下:

#include<stdio.h>
#include<math.h>
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
            break;
        for(int i=sqrt(2*m);i>=1;i--)
        {
            int na1=m- (i*(i-1)/2);
            if( na1%i == 0 )
                printf("[%d,%d]\n",na1/i,na1/i+i-1);
        }
        printf("\n");
    }
}