CodeForces - 165B Burning Midnight Oil ——二分

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

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.

题意：一共有n行代码，第一天写v行，第二天写k/v下取整行，第三天写k/v^2下取整行，依此类推，给定n和k的值，问v最小是多少

思路：二分法，v最小是1，最大是n，不断二分看是否能达到n

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <cstring>
#include <queue>
#include <algorithm>
#define ll long long
#define max_ 50
using namespace std;
ll n,k;
bool judge(ll x)
{
ll sum=x,p=k;
while(x/p!=0)
{
sum+=x/p;
p=p*k;
}
if(sum>=n)
return true;
else
return false;
}
int main(int argc, char const *argv[]) {
scanf("%lld%lld",&n,&k);
ll l=1,r=n,ans;
while(l<=r)
{
ll mid=(l+r)>>1;
if(judge(mid))
{
ans=mid;
r=mid-1;
}
else
l=mid+1;
}
printf("%lld\n",ans );
return 0;
}