A. Robot Sequence
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.

Input

The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.

The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.

Output

Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.

Examples
input
6
URLLDR
output
2
input
4
DLUU
output
0
input
7
RLRLRLR
output
12
Note

In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.

Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.

#include<bits/stdc++.h>
using namespace std;
int ud[202];
int rl[202];
int main()
{
int n;
char a; scanf("%d",&n);getchar();
for(int i=0;i<n;i++)
{
scanf("%c",&a);
if(a=='U') ud[i]=1;
else if(a=='D') ud[i]=-1;
else if(a=='R') rl[i]=1;
else rl[i]=-1; }
int ans=0;
int cntud=0;
int cntrl=0;
for(int i=0;i<n;i++)
{
cntud=ud[i];
cntrl=rl[i];
for(int j=i+1;j<n;j++)
{ cntud+=ud[j];
cntrl+=rl[j];
if(cntud==0&&cntrl==0) ans++;
}
}
printf("%d\n",ans);
return 0;
}

  

A. Robot Sequence的更多相关文章

  1. Codeforces 626A Robot Sequence

    A. Robot Sequence time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...

  2. Codeforces 626A Robot Sequence(模拟)

    A. Robot Sequence time limit per test:2 seconds memory limit per test:256 megabytes input:standard i ...

  3. 8VC Venture Cup 2016 - Elimination Round A. Robot Sequence 暴力

    A. Robot Sequence 题目连接: http://www.codeforces.com/contest/626/problem/A Description Calvin the robot ...

  4. Codeforces 626 A. Robot Sequence (8VC Venture Cup 2016-Elimination Round)

      A. Robot Sequence   time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  5. 前端自动化测试工具doh学习总结(二)

    一.robot简介 robot是dojo框架中用来进行前端自动化测试的工具,doh主要目的在于单元测试,而robot可以用来模仿用户操作来测试UI.总所周知,Selenium也是一款比较流行的前端自动 ...

  6. 8VC Venture Cup 2016 - Elimination Round

    在家补补题   模拟 A - Robot Sequence #include <bits/stdc++.h> char str[202]; void move(int &x, in ...

  7. Codeforces-8VC Venture Cup 2016-Elimination Round-626A.暴力 626B.水题 626C.二分

    A. Robot Sequence time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...

  8. Codeforces Round #389 Div.2 C. Santa Claus and Robot

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

  9. CSUFT 1002 Robot Navigation

    1002: Robot Navigation Time Limit: 1 Sec      Memory Limit: 128 MB Submit: 4      Solved: 2 Descript ...

随机推荐

  1. JS设置cookie,删除cookie

    js设置cookie有很多种方法. 第一种:(这个是w3c官网的代码) <script> //设置cookie function setCookie(cname, cvalue, exda ...

  2. linux命令合集

    序:介绍一些经常需要用到的Linux命令. 一.wget 作用:下载网络文件,将远程服务器文件恢复备份到本地. wget http://cn.wordpress.org/wordpress-3.1-z ...

  3. HDOJ 2063 过山车

    过山车 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submiss ...

  4. cf.295.B Two Buttons (bfs)

     Two Buttons time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  5. getVisibleSize 和 getContentSize 和 getWinSize

    getVisibleSize:获得视口(可视区域)的大小,若是DesignResolutionSize跟屏幕尺寸一样大,则getVisibleSize便是getWinSize.getVisibleOr ...

  6. Java笔记--泛型总结与详解

    泛型简介: 在泛型没有出来之前,编写存储对象的数据结构是很不方便的.如果要针对每类型的对象写一个数据结构,     则当需要将其应用到其他对象上时,还需要重写这个数据结构.如果使用了Object类型, ...

  7. SublimeText3 生成html标签快捷键

    mmet Documentation Syntax Child: > nav>ul>li <nav> <ul> <li></li> & ...

  8. iptables 命令介绍

    http://www.cnblogs.com/wangkangluo1/archive/2012/04/19/2457072.html iptables 防火墙可以用于创建过滤(filter)与NAT ...

  9. 《linux备份与恢复之二》3.19 dump(文件系统备份)

    <Linux指令从初学到精通>第3章文件管理,本章介绍了许多常用命令,如cp.ln.chmod.chown.diff.tar.mv等,因为这些都与文件管理相关,在日常的使用中经常用到,因此 ...

  10. 58. 分析、测试与总结:罗马数字和阿拉伯数字的转换[roman to integer and integer to roman in c++]

    [本文链接] http://www.cnblogs.com/hellogiser/p/roman-to-integer-and-integer-to-roman.html [题目] 给出一个罗马数字, ...