Codeforces 321 A-ciel and Robot

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" The main idea": A robot robot can act according to a given instruction, give you four kinds of instructions, robot initial position is (0,0), the instruction is out, robot will repeat the action, judge whether in the case of infinite action can pass the point (n,m).

" problem-solving ideas " in fact, carefully simulated, you can find that there is a cycle, judge can, see the Code bar ~ ~

Code:

#include <iostream> #include <algorithm> #include <bits/stdc++.h>using namespace std;const int n=105;    int N,m,c,t,ans,res,tmp;char str[n];int X[n],y[n];int main () {int x, y;        while (~SCANF ("%d%d", &x,&y)) {memset (x,0,sizeof (x));        memset (y,0,sizeof (Y));        scanf ("%s", str+1);        int Len=strlen (str+1);            if (x==0&&y==0) {puts ("Yes");        return 0;            } else{bool flag=0;            int j=0,xx=x;            int yy=y;            int uu=0,dd=0,ll=0,rr=0; for (int i=0, i<=len; ++i) {switch (Str[i]) {case ' U ': x[i]=x[i-1],y[i]                    =y[i-1]+1;///y[0]=1;                Break                    Case ' D ': x[i]=x[i-1],y[i]=y[i-1]-1;                Break                    Case ' L ': x[i]=x[i-1]-1,y[i]=y[i-1];                Break Case ' R ': x[i]=x[i-1]+1,y[i]=y[i-1];/X[0]=1, break;   }/* if (str[i]== ' U ') {x[i]=x[i-1],y[i]=y[i-1]+1; Uu++;///yy+=1} else if (str[i]== ' D ') {x[i]=x    [I-1],y[i]=y[i-1]-1; Dd++;///yy-=1} else if (str[i]== ' L ') {x[i]=x[    I-1]-1,Y[I]=Y[I-1]; Ll++;///xx-=1} else {X[i]=x[i-1]+1,y[i]=y[i    -1];            Xx+=1} */} int zq;                for (int i=0; i<=len; ++i) {//2 2 int za=xx-x[i];///2-x[0]=1, int zb=yy-y[i];///2-y[0]=1,                Zq= (X[len]!=0?za/x[len]:(y[len]!=0?zb/y[len]:1));         if (x[len]*zq==za&&y[len]*zq==zb&&zq>=0)///if (zq>=0) {puts ("Yes");            return 0;        }} puts ("No"); }} return 0;} /*2 2ru1 2ru-1 1000000000lrrlu0 0dyesnoyesyes*/

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Codeforces 321 A-ciel and Robot