Avito Cool Challenge 2018-A. Definite Game(思维题）

最新推荐文章于 2019-03-08 20:59:36 发布

转载最新推荐文章于 2019-03-08 20:59:36 发布 · 148 阅读

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原文链接：http://www.cnblogs.com/Staceyacm/p/10781916.html

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#数据结构与算法

本文介绍了一种基于数值操作的竞赛编程游戏，玩家通过选择特定的整数来减少初始值，目标是最小化最终结果。文章提供了游戏规则、示例输入输出及简洁的代码实现，展示了如何快速解决此类问题。

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time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Chouti was doing a competitive programming competition. However, after having all the problems accepted, he got bored and decided to invent some small games.

He came up with the following game. The player has a positive integer nn. Initially the value of nn equals to vv and the player is able to do the following operation as many times as the player want (possibly zero): choose a positive integer xx that x<nx<n and xx is not a divisor of nn, then subtract xx from nn. The goal of the player is to minimize the value of nn in the end.

Soon, Chouti found the game trivial. Can you also beat the game?

Input

The input contains only one integer in the first line: vv (1≤v≤1091≤v≤109), the initial value of nn.

Output

Output a single integer, the minimum value of nn the player can get.

Examples

input

Copy

output

Copy

input

Copy

output

Copy

Note

In the first example, the player can choose x=3x=3 in the first turn, then nn becomes 55. He can then choose x=4x=4 in the second turn to get n=1n=1 as the result. There are other ways to get this minimum. However, for example, he cannot choose x=2x=2 in the first turn because 22is a divisor of 88.

In the second example, since n=1n=1 initially, the player can do nothing.

题意：你可以选小于n的数x，并且这个数不能是n的约数，可以不选，然后n=n-x，然后在进行，直到最小的数。

题解：

我们可以通过直接选n-1，就得到了最小的1，但是要考虑2的情况，因为1是2的约数，故2的时候最小为2

代码非常简单

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
	int n;
	cin>>n;
	if(n==2)
    {
    	cout<<2<<endl;
	}
	else
	{
		cout<<1<<endl;
	}
	return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781916.html