Codeforces 322C Ciel and Robot【思维+模拟】细节很多= =

C. Ciel and Robot

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 string s.
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.

Examples

input

2 2
RU

output

Yes

input

1 2
RU

output

No

input

-1 1000000000
LRRLU

output

Yes

input

0 0
D

output

Yes

Note

In the first and second test case, command string is "RU", so the robot will go right, then go up, then right, and then up and so on.

The locations of its moves are (0, 0)  →  (1, 0)  →  (1,
1)  →  (2, 1)  →  (2,
2)  →  ...

So it can reach (2, 2) but not (1, 2).

题目大意：

按照给出的走路顺序无限重复下去，问能否走到点（a，b），起点在（0,0）；

思路：

细节很多。

①我们首先模拟多次重复，如果能够走到终点，那么就是Yes.

②否则我们记下模拟走一回合到达的位子xx，yy.如果此时一回合的走向将与目标点相反走向，结果就是No.如果此时xx==0&&yy==0，结果也是No
③除上述两种情况以外，我们再模拟一次走一回合的过程，当我们走到点（x，y）的时候，我们如果有：

x+K*xx==a&&y+K*yy==b，那么结果就是Yes.否则整个过程不能通过循环来得到，结果也是No.

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define ll __int64
char a[1500];
int main()
{
ll A,B;
while(~scanf("%I64d%I64d",&A,&B))
{
scanf("%s",a);
ll n=strlen(a);
ll xx,yy;
xx=yy=0;
int flag=0;
int tx=0,ty=0;
for(int i=0;i<100000;i++)
{
for(int j=0;j<n;j++)
{
if(tx==A&&ty==B)flag=1;
if(a[j]=='U')ty++;
if(a[j]=='D')ty--;
if(a[j]=='L')tx--;
if(a[j]=='R')tx++;
if(tx==A&&ty==B)flag=1;
}
}
for(ll i=0; i<n; i++)
{
if(xx==A&&yy==B)flag=1;
if(a[i]=='U')yy++;
if(a[i]=='D')yy--;
if(a[i]=='L')xx--;
if(a[i]=='R')xx++;
if(xx==A&&yy==B)flag=1;
}
if(flag==1)
{
printf("Yes\n");
continue;
}
if(xx==yy&&xx==0)
{
printf("No\n");
continue;
}
if(A*xx<0||B*yy<0)
{
printf("No\n");
continue;
}
ll x=0,y=0;
for(ll i=0; i<n; i++)
{
if((A-x)*yy==(B-y)*xx)
{
if(xx!=0&&(A-x)%xx==0)
flag=1;
if(yy!=0&&(B-y)%yy==0)
flag=1;
}
if(a[i]=='U')y++;
if(a[i]=='D')y--;
if(a[i]=='L')x--;
if(a[i]=='R')x++;
if((A-x)*yy==(B-y)*xx)
{
if(xx!=0&&(A-x)%xx==0)
flag=1;
if(yy!=0&&(B-y)%yy==0)
flag=1;
}
}
if(flag==1)printf("Yes\n");
else printf("No\n");
}
}