Burning Midnight Oil

Description

One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of n lines of code. Vasya is already exhausted, so he works like that: first he writes v lines of code, drinks a cup of tea, then he writes as much as  lines, drinks another cup of tea, then he writes  lines and so on: , , , ...

The expression  is regarded as the integral part from dividing number a by number b.

The moment the current value  equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished.

Vasya is wondering, what minimum allowable value v can take to let him write not less than n lines of code before he falls asleep.

Input

The input consists of two integers n and k, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1 ≤ n ≤ 10⁹, 2 ≤ k ≤ 10.

Output

Print the only integer — the minimum value of v that lets Vasya write the program in one night.

 利用二分搜索，属于找下界的题目，代码很简单，不过需要注意两种情况：
1. 当n<k的时候直接输出n即可，
2. 当n特别大的时候计算出来的sum值以及k^p的结果可能会超出int范围故需要用long long 保存。

代码如下：
#include<iostream>
using namespace std ;
int n ;
int k ;
bool judge(int v) ;
int main()
{
	while(cin>>n>>k)
  {
	int x = 1 ;
	int y = n ;
	int v ;
	while(x < y)
	{
		v = (x + y)/2 ;

		if(judge(v))
			y = v ;
		else
			x = v + 1 ;
	}
	cout<<x<<endl;
  }
	return 0 ;
}
bool judge(int v)
{
	long long  sum ;
	long long  x ;
	sum = 0 ;
	x = 1 ;
	while(v/x)
	{
		sum += v/x ;
		x = x * k ;
	}
	if(sum >= n)
		return 1 ;
	else
		return 0 ;
}