CodeForces 321A Ciel and Robot

题目链接：http://codeforces.com/problemset/problem/321/A

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).

思路：模拟一下，记录操作所能到达的所有点，以最后一个位置为一个循环。注意走向和坐标方向相反的情况。

附上AC代码：

#include 
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
const int maxn = 105;
char str[maxn];
int a, b;
vector > v;

int main(){
	#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	while (~scanf("%d%d", &a, &b)){
		getchar();
		fgets(str, maxn, stdin);
		int x=0, y=0;
		if (x==a && y==b){
			//printf("-1\n");
			printf("Yes\n");
			continue;
		}
		v.clear();
		bool ok = true;
		pair t;
		for (int i=0; str[i]!='\n'; ++i){
			//printf("%c\n", str[i]);
			if (str[i] == 'U')
				++y;
			else if (str[i] == 'D')
				--y;
			else if (str[i] == 'L')
				--x;
			else
				++x;
			if (x==a && y==b){
				//printf("0\n");
				printf("Yes\n");
				ok = false;
				break;
			}
			t.first = x;
			t.second = y;
			v.push_back(t);
		}
		if (ok){
			bool f1=false, f2=false;
			int n1=v[v.size()-1].first, n2=v[v.size()-1].second;
			if (n1 != 0)
				f1 = true;
			if (n2 != 0)
				f2 = true;
			for (int i=0; i0&&n1<=0) || (t2>0&&n2<=0) || (t1<=0&&n1>0) || (t2<=0&&n2>0))
					continue;
				if (f1){
					if (t1%n1 == 0){
						int k = t1/n1;
						if (f2 && t2%n2==0 && t2/n2==k){
							ok = false;
							//printf("1\n");
							break;
						}
						else if (!f2 && t2==0){
							ok = false;
							//printf("2\n");
							break;
						}
					}
				}
				else{
					if (t1 == 0){
						if (f2 && t2%n2==0){
							ok = false;
							//printf("3\n");
							break;
						}
						else if (!f2 && t2==0){
							ok = false;
							//printf("4\n");
							break;
						}
					}
				}
			}
			if (!ok)
				printf("Yes\n");
			else
				printf("No\n");
		}
	}
	return 0;
}