Buggy Robot （思维）

最新推荐文章于 2021-12-20 15:59:45 发布

Shuo Zhang

最新推荐文章于 2021-12-20 15:59:45 发布

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本文链接：https://blog.youkuaiyun.com/NSSWTT/article/details/79393496

本文介绍了一个关于机器人路径规划的问题——BuggyRobot。任务是计算机器人执行一系列移动指令后仍能返回初始位置的最大正确指令数。文章提供了详细的解题思路及代码实现。

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题目链接---Buggy Robot

Buggy Robot

Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0, 0). The robot can process commands. There are four types of commands it can perform:

U — move from the cell (x, y) to (x, y + 1);
D — move from (x, y) to (x, y - 1);
L — move from (x, y) to (x - 1, y);
R — move from (x, y) to (x + 1, y).

Ivan entered a sequence of n commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0, 0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!

Input

The first line contains one number n — the length of sequence of commands entered by Ivan (1 ≤ n ≤ 100).

The second line contains the sequence itself — a string consisting of n characters. Each character can be U, D, L or R.

Output

Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.

Example

Input

4
LDUR

Output

Input

5
RRRUU

Output

Input

6
LLRRRR

Output

题意：每组样例给出一组命令，机器人执行若干命令后回到了原点，求机器人执行的最多命令

题解：机器人只要执行的命令中，只要L和R，U和D成对出现就会回到原点

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
char str[105];
int main()
{
	int n;
	while(cin>>n)
	{
		cin>>str;
		int l=0,r=0,u=0,d=0;
		int len=strlen(str);
		for(int i=0;i<len;i++)
		{
			if(str[i]=='L')
			l++;
			if(str[i]=='R')
			r++;
			if(str[i]=='U')
			u++;
			if(str[i]=='D')
			d++; 
		}
		int ans=2*min(l,r)+2*min(u,d);
		cout<<ans<<endl; 
	}
return 0;
}