[补]CodeForces 165B Burning Midnight Oil

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#codeforces #题解 #c++

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探讨了在限定时间内完成程序编写的方法，通过不断降低效率的条件下，寻找完成特定行数代码所需的初始速度v的最小值。使用二分查找算法来确定这个最小的初始值。

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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 $\text{[math]}$ lines, drinks another cup of tea, then he writes $\text{[math]}$ lines and so on: $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ , ...

The expression $\text{[math]}$ is regarded as the integral part from dividing number a by number b.

The moment the current value $\text{[math]}$ 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
inputCopy
7 2
outputCopy
4
inputCopy
59 9
outputCopy
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;
}