B - Burning Midnight Oil CodeForces - 165B

最新推荐文章于 2020-07-20 10:11:29 发布

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本文介绍了一个有趣的编程挑战：Vasya要在一夜之间完成指定行数的编程任务，但他的工作效率会随时间逐渐降低。文章通过二分查找算法找到了满足条件的最小初始效率值v，并提供了完整的代码实现。

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.

Example
Input
7 2
Output
4
Input
59 9
Output
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.

#include<bits/stdc++.h>
using namespace std;
long long n,k,v,l,r,ans;
int find(long long num){
    long long tmp = num/k;
    v = num;
    if(v>=n) return 1;
    while(tmp!=0){
        v += tmp;
        tmp /= k;
        if(v>=n) return 1; 
    }
    return 0;
}
int main(){
    cin>>n>>k;
    r = n;
    l = 1;
    if(n<=k) ans = n;
    else{
        while(l<=r){
            long long mid;
            mid = (l+r)/2;
            if(find(mid)==1){
                ans = mid;
                r = mid-1;
            }
            else l = mid+1;
        }
    }
    cout<<ans<<endl;
    return 0;
}