Codeforces 991C 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 $n$ 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 $n$ ($1\le n\le {10}^{18}$) — 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

Copy

68

output

Copy

3

Note

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

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

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

题意：两个人吃糖，A每天吃k颗糖，B每天吃剩余糖的10%向下取整。A每天先吃，B后吃。问k至少为多少，能保证A总共吃的糖大于总量的一半。

思路：2分k+模拟检验k是否满足。

代码：.

#include 

using namespace std;

typedef long long ll;
ll n;

int check(ll x){
    ll tmp = n, tot = 0;
    while (tmp){
        if (tmp > x){
            tot += x;
            tmp -= x;
            tmp -= tmp/10;
        } else {
            tot += tmp;
            tmp = 0;
        }
    }
    if (tot >= n/2+(n%2))
        return 1;
    return 0;
}
int main()
{
    scanf("%I64d",&n);
    ll l = 1, r = n, mid;
    ll ans = n;
    while (r >= l){
        mid = (r+l)/2;
        int flag = check(mid);
        if (flag){
            r = mid-1;
            ans = min(ans,mid);
        } else {
            l = mid + 1;
        }
    }
    cout<