CF 321A(Ciel and Robot-暴力枚举)

大致思路：

机器人必然走n个循环+1段

所以只要暴力枚举段

设到第i步时机器人在(xi,yi)

则xt=t*xn+xi   yt=t*yn+yi

乱搞即可……

过了Present Test(好激动！数回Div2 只做2题的菜)

结果Final Wa 了……

//我是白痴……

经鉴定，是我68行把||打成了&&……

我的Div 1 啊!!!!!!!!

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (50000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
ll a,b;
int n;
char s[1000];
int x[1000],y[1000];
int main()
{
//  freopen("robot.in","r",stdin);
//  freopen(".out","w",stdout);
  cin>>a>>b;
  scanf("%s",s+1);n=strlen(s+1);
  x[0]=y[0]=0;
  bool bo=0;
  For(i,n)
  {
    if (s[i]=='R') x[i]=x[i-1]+1,y[i]=y[i-1];
    if (s[i]=='L') x[i]=x[i-1]-1,y[i]=y[i-1];
    if (s[i]=='U') x[i]=x[i-1],y[i]=y[i-1]+1;
    if (s[i]=='D') x[i]=x[i-1],y[i]=y[i-1]-1;
    if (x[i]==a&&y[i]==b) bo=1;
  }
  if (bo==1)
  {
    cout<<"Yes"<<endl;
    return 0;
  }
  Rep(i,n+1)
  {
    int t1=a-x[i],t2=b-y[i];
    if (t1!=0&&x[n]==0) continue;
    if (t2!=0&&y[n]==0) continue;
    if (x[n]==0||y[n]==0)
    {
      bool bo1=0,bo2=0;
      if (x[n]==0) bo1=1;
      if (y[n]==0) bo2=1;
      if (bo1&&!bo2) if (t2%y[n]==0&&t2/y[n]>=0) {cout<<"Yes"<<endl;return 0;} else continue; 
      if (!bo1&&bo2) if (t1%x[n]==0&&t1/x[n]>=0) {cout<<"Yes"<<endl;return 0;} else continue;
      if (!x[i]&&!y[i]) {cout<<"Yes"<<endl;return 0;} else continue;
    }
    if (t1%x[n]||t2%y[n]) continue;
    t1/=x[n],t2/=y[n];
    if (t1<0||t2<0||t1^t2) continue;
    cout<<"Yes"<<endl;
    return 0;
  }
  cout<<"No"<<endl;
  
  return 0;
}