codeforces A. Ciel and Robot

最新推荐文章于 2020-12-21 21:36:25 发布

淡定的小Y

最新推荐文章于 2020-12-21 21:36:25 发布

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本文链接：https://blog.youkuaiyun.com/xuezhongfenfei/article/details/9201623

本文探讨了一个关于机器人在二维平面上遵循指令移动并判断是否能到达指定坐标点的问题。通过解析输入字符串指令，模拟机器人移动过程，并判断最终位置是否与目标坐标匹配。

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Fox Ciel has a robot on a 2D plane. Initially it is located in (0, 0). Fox Ciel code a command to it. The command was represented by strings. Each character of s is one move operation. There are four move operations at all:

'U': go up, (x, y) → (x, y+1);
'D': go down, (x, y) → (x, y-1);
'L': go left, (x, y) → (x-1, y);
'R': go right, (x, y) → (x+1, y).

The robot will do the operations in s from left to right, and repeat it infinite times. Help Fox Ciel to determine if after some steps the robot will located in (a, b).

Input

The first line contains two integers a and b, ( - 10⁹ ≤ a, b ≤ 10⁹). The second line contains a string s (1 ≤ |s| ≤ 100, s only contains characters 'U', 'D', 'L', 'R') — the command.

Output

Print "Yes" if the robot will be located at (a, b), and "No" otherwise.

Sample test(s)

input
2 2
RU

output
Yes

input
1 2
RU

output
No

input
-1 1000000000
LRRLU

output
Yes

input
0 0
D

output
Yes

这道题的思路很简单，大家都可以想到，但是我写的程序bug太多了，昨晚没调出来，今天在测试数据下还调了几乎一上午，不得不说自己考虑什么真的很不全面，其实都是对除0，以及正负号的，自己的思路很不严密啊，下面贴出修改了n次的代码，o(╯□╰)o！

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<climits>
#include<map>
using namespace std;
#define rep(i,n) for(int i=0; i<n; i++)
#define repf(i,n,m) for(int i=(n); i<=(m); ++i)
#define repd(i,n,m) for(int i=(n); i>=(m); --i)
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
#define fab(a) ((a)>0?(a):(0-(a)))
#define ll long long
#define arc(a) ((a)*(a))
#define inf 100000
#define exp 0.000001
#define N 200
char s[N];
ll x,y;
struct node{
	ll x,y;
}a[N];
int main()
{ 
	int sign=0;
   while(cin>>x>>y)
	{
		sign++; 
		cin>>s;
        a[0].x=a[0].y=0;
		int len=strlen(s);
		int flag=0;
        rep(i,len)
		{
			a[i+1].x=a[i].x;a[i+1].y=a[i].y;
           if(s[i]=='U')  a[i+1].y++;
		   if(s[i]=='D')  a[i+1].y--;
		   if(s[i]=='L')  a[i+1].x--;
		   if(s[i]=='R')  a[i+1].x++;
		   if(a[i+1].x==x && a[i+1].y==y)
			   flag=1;
		} 
		if(x==0 && y==0)
		{printf("Yes\n"); continue;}
		ll xx=a[len].x,yy=a[len].y;
		if((xx==0 && yy==0)&& flag==0)
		{printf("No\n");continue;}  
        if(flag==0)
        { 
		  rep(i,len+1)
		  {  
             if(xx==0 && yy!=0 && (y-a[i].y)%yy==0&& a[i].x==x)
               flag=1;
             if(yy==0 && xx!=0 &&(x-a[i].x)%xx==0 && y==a[i].y)
              flag=1;
             if(xx*yy!=0 && (x-a[i].x)%xx==0 &&(y-a[i].y)%yy==0)
              flag=1;
             if(flag==1)
             {   
                 if(xx*yy!=0 && (x-a[i].x)/xx!=(y-a[i].y)/yy)
                  flag=0;
                 if(xx!=0 && (x-a[i].x)*xx<0)
                 flag=0;
                 if(yy!=0 && (y-a[i].y)*yy<0)
                 flag=0;
                 if(flag==1)
                   break;
             }
		  }
        }
		if(flag==1)
			printf("Yes\n");
		else
			printf("No\n");

	 }
   return 0;
}