Ciel and Robot CodeForces

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).
可以无限次的重复给出的过程，问可不可以达到目标点。想想都不可能是直接暴力的模拟。肯定有巧妙地方法。我们仔细想想，走到每一步的时候，在经过好几个轮回都有可能到达目标点。所以我们需要记录走每一步的时候坐标改变的程度，以及走完一个轮回坐标改变的程度。然后在遍历一遍就可以了。
代码如下：

#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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