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;
    }
}