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

最新推荐文章于 2021-03-21 17:56:49 发布

转载最新推荐文章于 2021-03-21 17:56:49 发布 · 114 阅读

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#数据结构与算法

本文解析了Codeforces上的一道A级题目，介绍了一个简单的算法问题：如何计算字符串中能够使机器人回到起始位置的有效子串数量。通过使用前缀和的方法实现了O(n^2)的解决方案。

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A. Robot Sequence

题目连接：

http://www.codeforces.com/contest/626/problem/A

Description

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.

Sample Input

6
URLLDR

Sample Output

Hint

题意

给你一个命令，然后让你输出这个命令中有多少个子串的起点和终点是一样的

题解：

数据范围只有200，所以直接n^2暴力就好了

记录一下前缀和。

其实不记录前缀和 n^3也是兹瓷的

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 503;
char s[maxn];
int x[maxn],y[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    scanf("%s",s+1);
    for(int i=1;i<=n;i++)
    {
        x[i]=x[i-1];
        y[i]=y[i-1];
        if(s[i]=='U')x[i]++;
        if(s[i]=='D')x[i]--;
        if(s[i]=='L')y[i]++;
        if(s[i]=='R')y[i]--;
    }
    int ans = 0;
    for(int i=0;i<=n;i++)
        for(int j=i+1;j<=n;j++)
            if(x[i]-x[j]==0&&y[i]-y[j]==0)
                ans++;
    cout<<ans<<endl;
}