CF991C Candies

原题链接：http://codeforces.com/contest/991/problem/C

Candies

After passing a test, Vasya got himself a box of $n$ candies. He decided to eat an equal amount of candies each morning until there are no more candies. However, Petya also noticed the box and decided to get some candies for himself.

This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$, same for all days. After that, in the morning he eats $k$ candies from the box (if there are less than $k$ candies in the box, he eats them all), then in the evening Petya eats $10\%$ of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats $k$ candies again, and Petya — $10\%$ of the candies left in a box, and so on.

If the amount of candies in the box is not divisible by $10$, Petya rounds the amount he takes from the box down. For example, if there were $97$ candies in the box, Petya would eat only $9$ of them. In particular, if there are less than
$10$ candies in a box, Petya won’t eat any at all.

Your task is to find out the minimal amount of $k$ that can be chosen by Vasya so that he would eat at least half of the $n$ candies he initially got. Note that the number $k$ must be integer.

Input

The first line contains a single integer $n$ $(1≤n≤10^{18})$ — the initial amount of candies in the box.

Output

Output a single integer — the minimal amount of $k$ that would allow Vasya to eat at least half of candies he got.

Example

input

68

output

3

Note

In the sample, the amount of candies, with $k=3$, would change in the following way (Vasya eats first):

$68→65→59→56→51→48→44→41→37→34→31→28→26→23→21→18→17→14→13→10→9→6→6→3→3→0$.

In total, Vasya would eat $39$ candies, while Petya — $29$.

题解

因为每次糖果都会缩水$10\%$，$10^{18}$只要$300$多次就没了，那么就愉快的二分答案吧，复杂度$O(log_2n\times log_{\frac{10}{9}}10^{18})$。

代码

#include<bits/stdc++.h>
using namespace std;
long long n,le,ri,mid;
bool check(long long k)
{
    long long less=0,res=n;
    for(;res>0;res-=k,less+=(res>0?res:0)/10,res-=res/10);
    return less<=n/2;
}
void in(){scanf("%I64d",&n);}
void ac()
{
    le=1,ri=n/2+1;
    while(le^ri)mid=le+ri>>1,check(mid)?ri=mid:le=mid+1;
    printf("%I64d",le);
}
int main(){in();ac();}