Educational Codeforces Round 32 B题 Buggy Robot（模拟）

B. Buggy Robot

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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

题意：每组样例给一个 命令串 ， 机器人在执行命令串时 会忽略一些 命令 并回到 起点 ， 问机器人最多执行了 多少命令

思路：分别取 向左向右 命令的最小数量和向上向下 命令的最小数量 ， 然后累加 乘 2

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

using namespace std ;

#define maxn 10000
char str[maxn] ;
int num[maxn] ;
int n ;

int main() {
while(~scanf(" %d" , &n  )) {

memset(num , 0 , sizeof(num)) ;

for(int i=1 ; i<=n ; i++) {
scanf(" %c",&str[i]) ;
}

for(int i=1 ; i<= n ; i++) {
if(str[i] == 'U') {
num[1] ++ ;
} else if(str[i] == 'D') {
num[2] ++ ;
} else if(str[i] == 'L') {
num[3] ++ ;
} else if(str[i] == 'R') {
num[4] ++ ;
}
}
int result = 0 ;
result += min(num[1] , num[2]) ;
result += min(num[3] , num[4]) ;

printf("%d\n" , result * 2  ) ;
}

return 0 ;
}