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).
可以無限次的重復給出的過程，問可不可以達到目標點。想想都不可能是直接暴力的模擬。肯定有巧妙地方法。我們仔細想想，走到每一步的時候，在經過好幾個輪回都有可能到達目標點。所以我們需要記錄走每一步的時候坐標改變的程度，以及走完一個輪回坐標改變的程度。然后在遍歷一遍就可以了。
代碼如下：

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std;const int maxx=101; char s[maxx]; int x,y; int a[maxx]; int b[maxx];int main() {while(scanf("%d%d",&x,&y)!=EOF){int xx=0,yy=0;getchar();gets(s);if(xx==x&&yy==y) {cout<<"Yes"<<endl;continue;}int flag=0;int len=strlen(s);for(int i=0;i<len;i++){if(s[i]=='U') yy++;if(s[i]=='D') yy--;if(s[i]=='R') xx++;if(s[i]=='L') xx--;a[i]=xx;b[i]=yy;}int k; for(int i=0;i<len;i++){if(xx){k=(x-a[i])/xx;}else if(yy){k=(y-b[i])/yy;}if(k<0) k=0;if(x==k*xx+a[i]&&y==k*yy+b[i]){flag=1;break;}}if(flag) cout<<"Yes"<<endl;else cout<<"No"<<endl;} }

努力加油a啊，(^o)/~

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