Description

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.

Sample Input

Input
RRU
Output
-1
Input
UDUR
Output
1
Input
RUUR
Output
2

Hint

In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.

In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change s to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.

题解:水题。

代码如下:

 #include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e 2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std; char a[];
int b[];
int main()
{
cin>>a;
int l=strlen(a);
if(l%)
{
cout<<"-1"<<endl;
return ;
}
for(int i=;i<l;i++)
{
if(a[i]=='L')
b[]++;
if(a[i]=='R')
b[]--;
if(a[i]=='U')
b[]++;
if(a[i]=='D')
b[]--;
}
//cout<<b[0]<<" "<<b[1]<<endl;
cout<<(abs(b[])+abs(b[]))/<<endl;
return ;
}

Memory and Trident(CodeForces 712B)的更多相关文章

  1. (CodeForces - 5C)Longest Regular Bracket Sequence(dp+栈)(最长连续括号模板)

    (CodeForces - 5C)Longest Regular Bracket Sequence time limit per test:2 seconds memory limit per tes ...

  2. Sorted Adjacent Differences(CodeForces - 1339B)【思维+贪心】

    B - Sorted Adjacent Differences(CodeForces - 1339B) 题目链接 算法 思维+贪心 时间复杂度O(nlogn) 1.这道题的题意主要就是让你对一个数组进 ...

  3. (CodeForces 558C) CodeForces 558C

    题目链接:http://codeforces.com/problemset/problem/558/C 题意:给出n个数,让你通过下面两种操作,把它们转换为同一个数.求最少的操作数. 1.ai = a ...

  4. [题解]Yet Another Subarray Problem-DP 、思维(codeforces 1197D)

    题目链接:https://codeforces.com/problemset/problem/1197/D 题意: 给你一个序列,求一个子序列 a[l]~a[r] 使得该子序列的 sum(l,r)-k ...

  5. 【Codeforces】【图论】【数量】【哈密顿路径】Fake bullions (CodeForces - 804F)

    题意 有n个黑帮(gang),每个黑帮有siz[i]个人,黑帮与黑帮之间有有向边,并形成了一个竞赛完全图(即去除方向后正好为一个无向完全图).在很多年前,有一些人参与了一次大型抢劫,参与抢劫的人都获得 ...

  6. Maximum Sum of Digits(CodeForces 1060B)

    Description You are given a positive integer nn. Let S(x) be sum of digits in base 10 representation ...

  7. 【日常训练】Help Victoria the Wise(Codeforces 99C)

    题意与分析 这题意思是这样的:在正方体的六面镶嵌给定颜色的宝石(相同颜色不区分),然后问最多有几种彼此不等价(即各种旋转过后看起来一致)的方案. 其实可以乱搞,因为范围只有720.求出全排列,然后每个 ...

  8. 【日常训练】Help Far Away Kingdom(Codeforces 99A)

    题意与分析 题意很简单,但是注意到小数可能有一千位,作为一周java选手的我选择了java解决. 这里的分析会归纳一些必要的Java API:(待补) 代码 /* * ACM Code => c ...

  9. Palindrome Degree(CodeForces 7D)—— hash求回文

    学了kmp之后又学了hash来搞字符串.这东西很巧妙,且听娓娓道来. 这题的题意是:一个字符串如果是回文的,那么k值加1,如果前一半的串也是回文,k值再加1,以此类推,算出其k值.打个比方abaaba ...

随机推荐

  1. Excel 如何判断某列哪些单元格包含某些字符

    “条件格式”,公式: =IF(COUNTIF($A2,,,) 然后根据需要设置格式

  2. ADO.Net的发展史

    1.演变历史: 它们是按照这个时间先后的顺序逐步出现的,史前->ODBC->OLEDB->ADO->ADO.Net. 2.下面分别介绍一下这几个. a. 史前的数据访问是什么样 ...

  3. iou与giou对比

    设矩形1大小为100x100,矩形2从左上角顶点重合开始,向右滑动250个单位. c++源码(基于opencv3.4.0) float iou(const cv::Rect& r1, cons ...

  4. HDU 1024 Max Sum Plus Plus【DP】

    Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we ...

  5. Python 使用有道翻译

    最近想将一些句子翻译成不同的语言,最开始想使用Python向有道发送请求包的方式进行翻译. 这种翻译方式可行,不过只能翻译默认语言,不能选定语言,于是我研究了一下如何构造请求参数,其中有两个参数最复杂 ...

  6. dao层、service和action的运用和区别

    DAO层叫数据访问层,全称为data access object,属于一种比较底层,比较基础的操作,对于数据库的操作,具体到对于某个表的增删改查, 也就是说某个DAO一定是和数据库的某一张表一一对应的 ...

  7. python笔记--异常处理

    Python异常处理 常见异常 AttributeError:属性错误,特性引用和赋值失败时会引发属性错误 NameError:试图访问的变量名不存在 SyntaxError:语法错误,代码形式错误 ...

  8. 记flask连接容联云时提示172001,网络错误

    直接用sms.py发送没有问题,直接从写好的注册页面发送就不行.在网上查了不少方法,试过了依然没用,结果换了一个网络就好了,估计是部分网络无法正常发送..后来问了下是环境问题,开发环境不稳定

  9. Generator

    基本概念 Generator函数是ES6提供的一种异步编程解决办法,语法行为与传统函数完全不同. Generator函数有多种理解角度.语法上,首先可以把它理解成,Generator函数是一个状态机, ...

  10. vue数据变动监测

    原文链接:https://blog.csdn.net/man_tutu/article/details/72148362 对象: 不能监测到: var vm = new Vue({ data:{ a: ...