Yukari's Birthday

Description

Today is Yukari's n-th birthday. Ran and Chen hold a celebration party for her. Now comes the most important part, birthday cake! But it's a big challenge for them to place n candles on the top of the cake. As Yukari has lived for such a long long time. Though she herself insists that she is a 17-year-old girl.

To make the birthday cake look more beautiful, Ran and Chen decide to place them like r ≥ 1 concentric circles. They place kⁱ candles equidistantly on the i-th circle, where k ≥ 2, 1 ≤ i ≤ r. And it's optional to place at most one candle at the center of the cake. In case that there are a lot of different pairs of r and k satisfying these restrictions, they want to minimize r × k. If there is still a tie, minimize r.

Input

There are about 10,000 test cases. Process to the end of file.

Each test consists of only an integer 18 ≤ n ≤ 10¹².

Output

For each test case, output r and k.

Sample Input

18

111

1111

Sample Output

1 17

2 10

3 10

 

 

 

枚举+二分

刚开始还以为是线性规划……到网上搜了下才知道是枚举+二分

 

 

/*

pow(int, double) 用于long long 的时候可能会出错，比如爆掉。

所以要重新写一个计算long long 的pow函数

*/

#include <iostream>

#include <cmath>

using namespace std;



long long powForLL(long long a, int b)    //long long 的pow函数

{

    long long ans = 1;

    

    for (int i = 0; i < b; ++i)

        ans *= a;

    

    return ans;

}



int main()

{

    long long n, r, k;

    

    while (cin >> n)

    {

        r = 1, k = n-1;

        

        for (int i = 2; i <= 45; ++i) //why 45?

        {

            long long ll = 2, rr = pow(n, 1.0/i), mm;    // ll rr mm都是关于k的 i是关于r的 

            

            while (ll <= rr)

            {

                mm = (long long)(ll+rr)/2;

                

                long long ans = (mm-powForLL(mm,(double)i+1))/(1-mm);//等比数列求和公式别说你不懂 

                

                if (ans == n || ans == n-1)    //中间不插蜡烛为n,插就n-1 

                {

                    if (mm*i < r*k)

                    {

                        r = i, k = mm;

                    }

                    break;

                }

                else if (ans > n)

                {

                    rr = mm-1;

                }

                else

                {

                    ll = mm+1;

                }

            }

        }

        

        cout << r << ' ' << k << endl;

    }

}

View Code