Educational Codeforces Round 32 B. Buggy Robot（模拟）

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本文介绍了一种针对机器人路径指令的校正算法，通过计算不同方向移动指令的配对数量来确定机器人返回起始位置的最大可行步骤。该算法应用于解决机器人在无限网格中根据一系列指令移动后能否回到原点的问题。

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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.

Examples

input
4
LDUR

output
4

input
5
RRRUU

output
0

input
6
LLRRRR

output
4

题解：

这题还是比较水的，直接记录下四个方向执行了几次然后取相反方向的最小值乘2就是答案

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
#define lson k*2
#define rson k*2+1
#define M (t[k].l+t[k].r)/2
#define ll long long
int main()
{
    int n,i,j;
    int x=0,y=0,ans=0;
    char s[105];
    int a[5];
    memset(a,0,sizeof(a));
    scanf("%d%s",&n,s);
    for(i=0;i<strlen(s);i++)
    {
        switch(s[i])
        {
            case 'L':a[0]++;break;
            case 'R':a[1]++;break;
            case 'U':a[2]++;break;
            case 'D':a[3]++;break;
        }
    }
    if(a[0]!=0&&a[1]!=0)
    {
        ans+=min(a[0],a[1])*2;
    }
    if(a[2]!=0&&a[3]!=0)
    {
        ans+=min(a[2],a[3])*2;
    }
    printf("%d\n",ans);
	return 0;
}