Memory and Trident(CodeForces 712B)
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
RRU
-1
UDUR
1
RUUR
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
- (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 ...
- 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 ...
- (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 ...
- [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 ...
- 【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 ...
- 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 ...
- 【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 ...
- 【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 ...
- 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
- Thinking in work
Scheduler? Realtime? sure SCI? Power supply and ECU life. how to assure? EMC?
- Vue PC后台系统组件大全
1.https://vue.ant.design/ 2.http://element-cn.eleme.io/#/zh-CN 3.https://www.iviewui.com/ 4.https:// ...
- MySQL主从复制虽好,能完美解决数据库单点问题吗?
一.单个数据库服务器的缺点 数据库服务器存在单点问题: 数据库服务器资源无法满足增长的读写请求: 高峰时数据库连接数经常超过上限. 二.如何解决单点问题 增加额外的数据库服务器,组建数据库集群: 同一 ...
- kubernets event 分析
1. event 是一个很重要的组成部分 event 分析 Kubernetes(K8s)Events介绍(上) Kubernetes Events介绍(中) Kubernetes Events介绍( ...
- java枚举类型详解
枚举类型是JDK1.5的新特性.显然,enum很像特殊的class,实际上enum声明定义的类型就是一个类.而这些类都是类库中Enum类的子类(java.lang.Enum<E>).它 ...
- java开发工具STS的下载及安装
将下载后的压缩文件解压,在解压后的sts-bundle下的sts-3.9.1RELEASE目录中STS.exe便是可执行程序,用于启动STS,将该文件发送到桌面快捷方式,当我们想使用STS时可以快速的 ...
- canvas与svg
canvas与svg都是用于在网页上绘制图形(位图). canvas是HTML5新出来的一个标签,用来定义一块画图的区域(canvas本身没有绘制能力),用JavaScript来画图,可以绘制路径.矩 ...
- 让selenium中的Cromerderive不加载图片设置
把配置参数(chrom_opt)设置好后将其添加到 browser = webdriver.Chrome(executable_path="chromedriver.exe的路径" ...
- KMP总结
首先给一个我能看懂的KMP讲解: http://blog.csdn.net/v_july_v/article/details/7041827 来自大神july 文章很长,但是慢慢看,会发现讲的很好. ...
- 《内蒙古自治区第十三届大学生程序设计竞赛试题_H 公孙玉龙》
这个题有点小坑,最坑的地方就是 输入的b 变量 有可能 是 负数 ! 负数 ! 负数! 对 ,你没有看错,就是负数,坑死我了, 一直都是 content.charAt(0) 判断 ...