codeforces 712B B. Memory and Trident(水题)

题目链接：

B. Memory and Trident

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string s with his directions for motion:

• An 'L' indicates he should move one unit left.
• An 'R' indicates he should move one unit right.
• A 'U' indicates he should move one unit up.
• A 'D' indicates he should move one unit down.

But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in s with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.

Input

The first and only line contains the string s (1 ≤ |s| ≤ 100 000) — the instructions Memory is given.

Output

If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.

Examples

input

RRU

output

-1

input

UDUR

output

1

input

RUUR

output

2

题意：

问最少改变多少个才能最后回到原点;

思路：

水水水;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
  
using namespace std;
  
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
  
typedef  long long LL;
  
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
  
const int mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=(1<<20)+10;
const int maxn=1e5+110;
const double eps=1e-12;
 

char s[maxn];
int main()
{
    scanf("%s",s);
    int len=strlen(s),l=0,r=0,u=0,d=0;
    for(int i=0;i<len;i++)
    {
        if(s[i]=='R')r++;
        else if(s[i]=='L')l++;
        else if(s[i]=='U')u++;
        else d++;
    }
    if(len&1)cout<<"-1
";
    else cout<<(abs(l-r)+abs(u-d))/2<<endl;
    return 0;
}