## C. Candies（二分）

C. Candies
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

After passing a test, Vasya got himself a box of nn 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 kk, same for all days. After that, in the morning he eats kk candies from the box (if there are less than kk candies in the box, he eats them all), then in the evening Petya eats 10%10% of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats kk candies again, and Petya — 10%10% of the candies left in a box, and so on.

If the amount of candies in the box is not divisible by 1010, Petya rounds the amount he takes from the box down. For example, if there were 9797 candies in the box, Petya would eat only 99 of them. In particular, if there are less than 1010 candies in a box, Petya won't eat any at all.

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

Input

The first line contains a single integer nn (1n10181≤n≤1018) — the initial amount of candies in the box.

Output

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

Example
input
Copy
68

output
Copy
3

Note

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

686559565148444137343128262321181714131096633068→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 3939 candies, while Petya — 2929.

#include "iostream"
#include "algorithm"
using  namespace std;
typedef long long ll;
int main()
{
ll n,l,r,mid,x,tmp,cnt,sum,ans=1e18;
cin>>n;
l=0,r=n;
while(l<=r){
sum=0;
mid=(l+r)>>1;
if(mid==0) {l+=1;continue;}//如果枚举出0，那么肯定是不可以的，所以l+1
tmp=n;
while(1){
if(tmp>=mid) tmp-=mid,sum+=mid;//如果剩余的大于mid，那么减去mid，否则，直接取完
else sum+=tmp,tmp=0;
tmp-=tmp/10;
x=n%2?(n/2)+1:n/2;//n/2不是整数，那么加一（相当与向上取整）
if(sum>=x) break;//如果a取的蛋糕比n的一半还要多，那么这个mid是可以的
if(tmp==0) {sum=-1;break;}
}
if(sum!=-1) r=mid-1,ans=min(ans,mid);//如果mid满足条件，更新ans
else l=mid+1;
}
cout<<ans<<endl;
return 0;
}