CF991C Candies 二分 第十五

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 (1≤n≤10181≤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):

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→068→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.

题意： 给你一个n，每次Vasya可以使数字减k，Petya可以使数字减少百分之十，问要使Vasya减去的数占一半或以上k最小是啥

k越大Vasya吃的越多，所以我们可以把1-n看作一个排好序的结果集。

在有序的集合中找最小结果可以直接二分模拟

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + ;
const int mod = 1e9 + ;
typedef long long ll;
ll n;
ll check( ll x ) {
    ll now, sum;
    now = n, sum = ;
    while( now ) {
        if( now <= x ) {
            sum += now;
            break;
        }
        now -= x;
        sum += x;
        now -= now / ;
    }
    return sum;
}
int main(){
    std::ios::sync_with_stdio(false);
    while( cin >> n ) {
        ll le = , ri = n, mid;
        while( le < ri ) {
            mid = ( le + ri ) / ;
            if( check(mid) >= ( n +  ) /  ) {
                ri = mid;
            } else {
                le = mid + ;
            }
        }
        cout << le << endl;
    }
    return ;
}