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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
示例1
输入
A10;S20;W10;D30;X;A1A;B10A11;;A10;
输出
10,-10
代码1:
//第十七题 坐标移动
#include <iostream>
#include <string>
#include <cstddef>
#include<vector>
using namespace std;
int main()
{
string inStr;
while (cin >> inStr)
{
int iPosition1 = 0;
int iPosition2 = 0;
size_t fPosition = 0;
vector<string>tMove;
while ((fPosition = inStr.find(';')) != string::npos)
{
string temp(inStr.substr(0, fPosition));
if (temp.length() > 3 || temp.length() < 2)
{
inStr.erase(0, fPosition + 1);
continue;
}
tMove.push_back(temp);
inStr.erase(0, fPosition + 1);
}
int iMax = tMove.size();
for (int i = 0; i < iMax; i++)
{
string dStr = tMove.at(i);
int iLength = dStr.length();
switch (dStr.at(0))
{
case 'A':
{
int iNum = 0;
for (int j = 1; j < iLength; j++)
{
if (isdigit(dStr.at(j)))
{
iNum = iNum * 10 + dStr.at(j) - '0';
}
else
{
iNum = 0;
break;
}
}
iPosition1 -= iNum;
}
break;
case 'D':
{
int iNum = 0;
for (int j = 1; j < iLength; j++)
{
if (isdigit(dStr.at(j)))
{
iNum = iNum * 10 + dStr.at(j) - '0';
}
else
{
iNum = 0;
break;
}
}
iPosition1 += iNum;
}
break;
case 'S':
{
int iNum = 0;
for (int j = 1; j < iLength; j++)
{
if (isdigit(dStr.at(j)))
{
iNum = iNum * 10 + dStr.at(j) - '0';
}
else
{
iNum = 0;
break;
}
}
iPosition2 -= iNum;
}
break;
case 'W':
{
int iNum = 0;
for (int j = 1; j < iLength; j++)
{
if (isdigit(dStr.at(j)))
{
iNum = iNum * 10 + dStr.at(j) - '0';
}
else
{
iNum = 0;
break;
}
}
iPosition2 += iNum;
}
break;
default:
break;
}
}
cout << iPosition1 << "," << iPosition2 << endl;
}
return 0;
}
代码2:
#include <iostream>
#include <string>
#include <cstddef>
using namespace std;
int main()
{
string str;
while (cin >> str)
{
pair<int, int> point(0, 0);
size_t found = str.find_first_of(';');
int start = 0;
while (found != str.npos)
{
string s1 = str.substr(start, found - start);
start = found + 1;
found = str.find_first_of(';', found + 1);
if (s1.size() >= 2 && s1.size() <= 3)
{
char c = s1[0];
int invalid = 0;
int sum = 0;
for (int i = 1; i < s1.size(); i++)
{
if (s1[i] >= '0' && s1[i] <= '9')
{
sum = sum * 10 + (s1[i] - '0');
}
else
{
invalid = 1;
break;
}
}
if (invalid == 0)
{
switch (c)
{
case 'A':
{
point.first -= sum;
break;
}
case 'D':
{
point.first += sum;
break;
}
case 'W':
{
point.second += sum;
break;
}
case 'S':
{
point.second -= sum;
break;
}
}
}
}
}
cout << point.first << "," << point.second << endl;
}
return 0;
}
测试的时候本人写的代码一测试时间5毫秒。代码二7毫秒