Codeforces Round #491 (Div. 2)：C. Candies（二分）

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Jaihk662

最新推荐文章于 2022-06-07 22:05:29 发布

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本文介绍了一个有趣的算法问题：如何确定Vasya每天从糖果盒中取出的最小数量的糖果，以便他能够消耗至少一半的糖果总量，同时考虑到Petya会吃掉剩余糖果的10%。文章提供了一种通过二分查找来解决该问题的有效方法。

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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 $n$ ( $1 \leq n \leq 10^{18}$ ) — 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

题意：有一盒糖罐，里面有n个糖果，Vasya和Petya轮流吃，Vasya先吃且每次固定吃x个（不足x个全吃完），Petya每次吃掉当前的10%（向下取整，可能为0），求最小的x使得Vasya能吃到至少一半的糖果

思路：二分答案，直接模拟检测

#include<stdio.h>
#define LL long long
LL n;
LL Jud(LL x)
{
	LL now, sum;
	now = n, sum = 0;
	while(now)
	{
		if(now<=x)
		{
			sum += now;
			break;
		}
		now -= x;
		sum += x;
		now -= now/10;
	}
	return sum;
}
int main(void)
{
	LL l, r, m;
	scanf("%I64d", &n);
	l = 1, r = n;
	while(l<r)
	{
		m = (l+r)/2;
		if(Jud(m)>=(n+1)/2)
			r = m;
		else
			l = m+1;
	}
	printf("%I64d\n", l);
	return 0;
}