HDU 4430 Yukari's Birthday二分查找

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
代码：

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std ;
 long long ab(long long a,int b){

    long long res=1;

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

    res*=a;

    return res;
    }
int main( )
{
	long long  n , m , k , mm ;
	long long  r , rr , ll , e ;
	long long ans ;
	int i ;
	while( scanf( "%lld" , &n ) != EOF )
	{
		r = 1 ;
		k = n - 1 ;
		// 因为k >= 2 所以半径不会很大
		for(  i = 2 ; i <= 45  ; i++)
		{
			//二分查找k值
			ll = 2 ;
			rr = (long long) pow( n , 1.0 / i ) ;
			//k最大是根号n
			while( ll <= rr )
			{
				mm =  ( long long )( ll + rr ) / 2  ;
				// 等比数列求和公式
				ans = ( mm - ab( mm , i + 1 ) ) / ( 1 - mm ) ;

				if( ans == n || n - 1 == ans )
				{   
				//cout << ans << endl ;
					if( r * k > i * mm )
					{
						r = i ;
						k = mm ;
					}
					break ;
				}
				else if( ans > n )
				{
					rr = mm - 1 ;
				}
				else ll = mm + 1 ;
			}
		}
		cout << r << " " << k << endl ;
	}
}