Codeforces Round #190 (Div. 2)-C. Ciel and Robot

原题链接

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 2RU

output

Yes

input

1 2RU

output

No

input

-1 1000000000LRRLU

output

Yes

input

0 0D

output

Yes

把robot沿着s串走一次称为一周期，一周期后走到(x, y)若x != 0 || y != 0则接下来每一周期结束后走到的点都在(0, 0)->(x, y)这条射线上.把(a, b)沿着s串逆方向走，观察是否在这条射线上

#include <bits/stdc++.h>#define maxn 105using namespace std;typedef long long ll;char s[maxn];int a, b, x, y, p1, p2;bool judge(int k1, int k2){if(k1 == 0 && k2 == 0) return true;if(x == 0 && y == 0){if(k1 == 0 && k2 == 0) return true;return false;}if(x == 0){if(k1 == 0 && (ll)y * k2 > 0 && k2 % y == 0)   return true;return false;}if(y == 0){if(k2 == 0 && (ll)x * k1 > 0 && k1 % x == 0)   return true;return false;}if((ll)x * k2 == (ll)y * k1 && (ll)x * k1 > 0 && (ll)y * k2 > 0 && k1 % x == 0 && k2 % y == 0) return true;return false;}int main(){//freopen("in.txt", "r", stdin);scanf("%d%d%s", &a, &b, s);x = 0;y = 0;for(int i = 0; s[i]; i++){    if(s[i] == 'L')     x--;    else if(s[i] == 'R')     x++;    else if(s[i] == 'D')     y--;    else     y++;}if(judge(a, b)){puts("Yes");return 0;}int len = strlen(s);for(int i = len-1; i >= 0; i--){p1 = a, p2 = b;for(int j = i; j >= 0; j--){if(s[j] == 'L')          p1++;    else if(s[j] == 'R')       p1--;    else if(s[j] == 'D')       p2++;        else          p2--;}if(judge(p1, p2)){puts("Yes");return 0;}}puts("No");return 0;}