zoukankan      html  css  js  c++  java
  • Ciel and Robot

    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| ≤ 100s only contains characters 'U', 'D', 'L', 'R') — the command.

    Output

    Print "Yes" if the robot will be located at (a, b), and "No" otherwise.

    细节蛮多的,做的我好忧伤。。。

     1 #include <iostream>
     2 #include <string>
     3 #include <map>
     4 #include <cstdio>
     5 #include <cmath>
     6 #include <cstring>
     7 using namespace std;
     8 
     9 int main()
    10 {
    11     char s[101];
    12     int a, b, dx, dy, i;
    13     while(scanf("%d %d", &a, &b) != EOF)
    14     {
    15         scanf("%s", s);
    16         dx = dy = 0;
    17         for(i = 0; s[i] != ''; i++)
    18         {
    19             if(dx == a && dy == b) break;
    20             if(s[i] == 'U') dy++;
    21             else if(s[i] == 'D') dy--;
    22             else if(s[i] == 'L') dx--;
    23             else dx++;
    24         }
    25         if(s[i] == '')
    26         {
    27             int dx2 = abs(dx), dy2 = abs(dy);
    28             for(i = 0; s[i] != ''; i++)
    29             {
    30                 if(s[i] == 'U') b--;
    31                 else if(s[i] == 'D') b++;
    32                 else if(s[i] == 'L') a++;
    33                 else a--;
    34                 if(!a && !b) break;
    35                 int a2 = abs(a), b2 = abs(b);
    36                 if((long long)a * dy == (long long)b * dx && (long long)b * dy >= 0 && (long long)a * dx >= 0)
    37                 {
    38                     if(dy && dx)
    39                     {
    40                         if(a2 % dx2 == 0 && b2 % dy2 == 0)  break;
    41                     }
    42                     else if(!dx && dy)
    43                     {
    44                         if(!a && b2 % dy2 == 0)  break;
    45                     }
    46                     else if(dx && !dy)
    47                     {
    48                         if(!b && a2 % dx2 == 0) break;
    49                     }
    50                 }
    51             }
    52         }
    53         if(s[i] != '') puts("Yes");
    54         else puts("No");
    55     }
    56     return 0;
    57 }
    View Code

  • 相关阅读:
    1203正规式转换为有穷自动机
    访问本班同学的博客
    0312复利计算3
    操作系统
    0311复利计算2
    0309简单的复利计算
    0302IT行业虽吃香,能完全享受这块“香"的也很难
    12.30递归下降语义分析
    1203正规式转换为有穷自动机
    对10位同学的文法解释和语法树的评论
  • 原文地址:https://www.cnblogs.com/cszlg/p/3242265.html
Copyright © 2011-2022 走看看