• 华为机试题 坐标移动


    题目描述

    开发一个坐标计算工具, A表示向左移动,D表示向右移动,W表示向上移动,S表示向下移动。从(0,0)点开始移动,从输入字符串里面读取一些坐标,并将最终输入结果输出到输出文件里面。

     

    输入:

     

    合法坐标为A(或者D或者W或者S) + 数字(两位以内)

     

    坐标之间以;分隔。

     

    非法坐标点需要进行丢弃。如AA10;  A1A;  $%$;  YAD; 等。

     

    下面是一个简单的例子 如:

     

    A10;S20;W10;D30;X;A1A;B10A11;;A10;

     

    处理过程:

     

    起点(0,0)

     

    +   A10   =  (-10,0)

     

    +   S20   =  (-10,-20)

     

    +   W10  =  (-10,-10)

     

    +   D30  =  (20,-10)

     

    +   x    =  无效

     

    +   A1A   =  无效

     

    +   B10A11   =  无效

     

    +  一个空 不影响

     

    +   A10  =  (10,-10)

     

     

     

    结果 (10, -10)



    输入描述:

    一行字符串



    输出描述:

    最终坐标,以,分隔


    输入例子:
    A10;S20;W10;D30;X;A1A;B10A11;;A10;

    输出例子:
    10,-10


    该题先预处理一下,取得所有的指令的列表,再一次处理会比较简单

    然后根据题意,正常的指令要么是3个字符A10   要么是2个字符  A1

    所以根据指令的长度先排除一些错误指令,然后根据指令的特点继续排除和处理指令

    这道题需要注意是要考虑处理多行指令,所以最外层要有while循环

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    #include<iostream>
    #include<vector>
     
    using namespace std;
     
    int main(){
        string s;
        while(getline(cin,s))
        {
            int heng=0;
            int zong=0;
            string temp;
            vector<string> vs;
            temp.clear();
            for(int i=0;i<s.size();i++){
                if(s[i]==';'){
                    vs.push_back(temp);
                    temp.clear();
                }
                else{
                    temp.push_back(s[i]);
                }
            }
            for(int i=0;i<vs.size();i++)
            {
                if(vs[i].size()==2 || vs[i].size()==3) 
                {
                    if(vs[i][0]=='A' || vs[i][0]=='W' || vs[i][0]=='S' || vs[i][0]=='D')
                    {
                        if(vs[i].size()==2){
                            if(vs[i][1]>='0' && vs[i][1]<='9')
                            {
                                if(vs[i][0]=='A')
                                    heng=heng-(vs[i][1]-'0');
                                else if(vs[i][0]=='D')
                                    heng=heng+(vs[i][1]-'0');
                                else if(vs[i][0]=='W')
                                    zong=zong+(vs[i][1]-'0');
                                else if(vs[i][0]=='S')
                                    zong=zong-(vs[i][1]-'0');  
                            }
                            else
                                continue;
                        }
                        else if(vs[i].size()==3){
                            if(vs[i][1]>='0' && vs[i][1]<='9' && vs[i][2]>='0' && vs[i][2]<='9')
                            {
                                if(vs[i][0]=='A')
                                    heng=heng-(vs[i][1]-'0')*10-(vs[i][2]-'0');
                                else if(vs[i][0]=='D')
                                    heng=heng+(vs[i][1]-'0')*10+(vs[i][2]-'0');
                                else if(vs[i][0]=='W')
                                    zong=zong+(vs[i][1]-'0')*10+(vs[i][2]-'0');
                                else if(vs[i][0]=='S')
                                    zong=zong-(vs[i][1]-'0')*10-(vs[i][2]-'0');
                            }
                            else
                                continue;
                        }
                    }
                    else
                        continue;
                }
                else
                    continue;
            }
            cout<<heng<<","<<zong<<endl;
            s.clear();
        }
    }





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  • 原文地址:https://www.cnblogs.com/gremount/p/5835626.html
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