[补]CodeForces 165B Burning Midnight Oil

B. Burning Midnight Oil

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

Copy

7 2

output

Copy

4

input

Copy

59 9

output

Copy

54

Note

In the first sample the answer is v = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task.

In the second sample the answer is v = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than n = 59.

题意：有n的任务需要完成，初始进度是v， 第一次加v/k , 第二次加 v/(k*k), 第三次加 v/(k*k*k), 求 完成进度的最小v

题解：对[0， n]进行二分 找到最小的 v值

AC代码：

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
//--------------------------
ll n, k, mid;

bool check() {
	ll ans = mid, t = k;
	if (ans >= n) return 1;
	while (mid >= t) {
		ans += mid / t;
		t *= k;
	}
	return ans >= n;
}

int main() {
	cin >> n >> k;
	ll l = 0, r = n, ans;
	while (l <= r) { //二分
		mid = l + r >> 1;
		if (check()) {
			r = mid - 1;	
			ans = mid;
		} else
			l = mid + 1;
	}
	cout << ans << endl;
	return 0;
}