Skyscraper

题目描述

Mr. Port plans to start a new business renting one or more floors of the new skyscraper with one giga floors, MinatoHarukas. He wants to rent as many vertically adjacent floors as possible, because he wants to show advertisement on as many vertically adjacent windows as possible. The rent for one floor is proportional to the floor number, that is, the rent per month for the n-th floor is n times that of the first floor. Here, the ground floor is called the first floor in the American style, and basement floors are out of consideration for the renting. In order to help Mr. Port, you should write a program that computes the vertically adjacent floors satisfying his requirement and whose total rental cost per month is exactly equal to his budget.

For example, when his budget is 15 units, with one unit being the rent of the first floor, there are four possible rent plans, 1+2+3+4+5, 4+5+6, 7+8, and 15. For all of them, the sums are equal to 15. Of course in this example the rent of maximal number of the floors is that of 1+2+3+4+5, that is, the rent from the first floor to the fifth floor.

输入

The input consists of multiple datasets, each in the following format.
b
A dataset consists of one line, the budget of Mr. Port b as multiples of the rent of the first floor. b is a positive integer satisfying 1 < b < 109.

The end of the input is indicated by a line containing a zero. The number of datasets does not exceed 1000.

输出

For each dataset, output a single line containing two positive integers representing the plan with the maximal number of vertically adjacent floors with its rent price exactly equal to the budget of Mr. Port. The first should be the lowest floor number and the second should be the number of floors.

样例输入

样例输出

1 5
16 1
2 1
1 2
16 4389
129 20995
4112949 206
15006 30011
46887 17718
163837 5994

原来是一个等差数列。把求和公式化简为是一个一元二次方程在利用韦达定理求解就行了。

（韦达定理：x1+x2= -(b/a) ; x1*x2=c/a）;

直接求是会超时的。一层循环是1e9，只要有另一层循环，不管是多小，一般是会超时的。

代码：

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
   int n;
   while(scanf("%d",&n)&&n!=0) //可以看出来定是整数的。
   {
       for(int i=(int)sqrt(2*n); i>=1 ;i--) //边界问题也是要注意。
       {
           double temp=(double)i;
           if( (2*n)/temp-(2*n)/i==0 ) //精度损失。
           {
               int flag=(2*n-i*i)/i;
               double abc=(double)i;
               if( (2*n-abc*abc)/abc - flag!=0 )
                   continue;
               int ans=(flag+1)/2;
               if( (flag+1)/2.0-ans!=0 )continue;
               cout<<ans<<" "<<i<<endl;
               break;
           }
       }
   }
   return 0;
}

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