E - Buggy Robot

最新推荐文章于 2025-08-31 15:34:28 发布

原创最新推荐文章于 2025-08-31 15:34:28 发布 · 315 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

45 篇文章

订阅专栏

本文介绍了一个关于机器人路径计算的问题，机器人初始位于坐标(0,0)，根据一系列指令移动，需要计算在忽略部分指令的情况下，机器人回到起点的最大指令数。文章提供了一种通过统计指令对数来解决问题的方法。

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：有一个机器人在点（0.0），会进行四种指令，机器人会选择性的忽略一些指令，问最多可能有多少个指令使机器人回到（0,0）；

思路：L与R相对，U与D相对，统计有多少对LR和UD，乘以2即可；

下面附上我的代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n,u=0,d=0,r=0,l=0;
	char s[105];
	cin>>n;
	cin>>s;
	for(int i=0;i<n;i++)
	{
		if(s[i]=='U') u++;
		if(s[i]=='D') d++;
		if(s[i]=='L') l++;
		if(s[i]=='R') r++;	
	}
	int ans1=min(l,r);
	int ans2=min(u,d);
	printf("%d\n",(ans1+ans2)*2);
	return 0;	
}