UVa272,UVa401,UVa10082

UVa272 Tex Quotes

用来熟悉输入输出的题，给一段文字，用getchar一个一个读并判断。

使用q作为标志，碰到第一个双引号，变换一下输出，q取反

碰到第二个双引号，变换输出，q取反

代码：

#include 

using namespace std;

int main()
{
    bool q = 0;
    char it;
    while((it=getchar())!=EOF)
    {
        if (it == '"')
        {
            if (q == 0)
                cout << "``";
            else
                cout << "''";
            q = !q;
        }
        else
        {
            cout << it;
        }
    }
    return 0;
}

UVa401 Palindromes

回文串与镜像串的判断
对于回文串的判断比较简单，字符串头和尾比较然后往中间收缩即可
我写了这个函数用来判断回文

bool isPalindromes(const string &judge)
{
    int i = 0;
    int len = (int)judge.size();
    while (i < ((len+1) / 2))
    {
        if (judge[i] == judge[judge.size() - i - 1])
        {
            i = i + 1;
        }
        else
        {
            return false;
        }
    }
    return true;
}

对于镜像串来说，比较巧妙的是用了常量字符数组，定义翻转函数，先列出A-Z和1-9的镜像字符，输入要判断的字符，做翻转，返回翻转后的字符然后与要判断的字符串相对应的位置比较

翻转函数：

char reverse(char judgeChar)
{
    const string alphaRev = "A   3  HIL JM O   2TUVWXY5";
    const string numRev = "1SE Z  8 ";
    if (isalpha(judgeChar))
    {
        return alphaRev[judgeChar - 'A'];
    }
    else
    {
        return numRev[judgeChar - '1'];
    }
}

判断镜像串的函数：

bool isMirrored(const string &judge)
{
    int i = 0;
    int len = (int)judge.size();
    while (i < ((len + 1) / 2))
    {
        if (reverse(judge[i]) == judge[judge.size() - i - 1])
        {
            i = i + 1;
        }
        else
        {
            /*if ((judge[i] == '0'&&judge[judge.size() - i - 1] == 'O') || (judge[i] == 'O'&&judge[judge.size() - i - 1] == '0'))
                i = i + 1;
            else*/
                return false;
        }
    }
    return true;
}

因为题目要求输出非回文，回文非镜像，镜像非回文，镜像回文四种情况，所以就做个判断，分别调用以上函数就好了

整体代码如下

#include 
#include 
#include 
using namespace std;

char reverse(char judgeChar)
{
    const string alphaRev = "A   3  HIL JM O   2TUVWXY5";
    const string numRev = "1SE Z  8 ";
    if (isalpha(judgeChar))
    {
        return alphaRev[judgeChar - 'A'];
    }
    else
    {
        return numRev[judgeChar - '1'];
    }
}

bool isPalindromes(const string &judge)
{
    int i = 0;
    int len = (int)judge.size();
    while (i < ((len+1) / 2))
    {
        if (judge[i] == judge[judge.size() - i - 1])
        {
            i = i + 1;
        }
        else
        {
            /*if ((judge[i] == '0'&&judge[judge.size() - i - 1] == 'O') || (judge[i] == 'O'&&judge[judge.size() - i - 1] == '0'))
                i = i + 1;
            else*/
            return false;
        }
    }
    return true;
}
bool isMirrored(const string &judge)
{
    int i = 0;
    int len = (int)judge.size();
    while (i < ((len + 1) / 2))
    {
        if (reverse(judge[i]) == judge[judge.size() - i - 1])
        {
            i = i + 1;
        }
        else
        {
                return false;
        }
    }
    return true;
}

void isPalindromesOutput(const string &output)
{
    cout << output << " -- is a regular palindrome."<void notPalindromesOutput(const string &output)
{
    cout << output << " -- is not a palindrome."<void isMirroredOutput(const string &output)
{
    cout << output << " -- is a mirrored string."<void isMirAndPalOutput(const string &output)
{
    cout << output << " -- is a mirrored palindrome."<int main()
{
    string needJudge;
    while (cin >> needJudge)
    {
        if (isPalindromes(needJudge))
        {
            if (isMirrored(needJudge))
            {
                isMirAndPalOutput(needJudge);
            }
            else
            {
                isPalindromesOutput(needJudge);
            }
        }
        else
        {
            if (isMirrored(needJudge))
            {
                isMirroredOutput(needJudge);
            }
            else
            {
                notPalindromesOutput(needJudge);
            }
        }
        cout << endl;
    }
    //system("pause");
    return 0;
}

UVA10082 WERTYU

这题也是常量字符的妙用，边输入，边判断，边输出，但要注意转义的问题，’\’字符特殊，需要变为‘\’
代码如下:

#include 

using namespace std;

const char s[] = "`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./";//注意转义问题
int main()
{
    char it;
    while ((it = getchar()) != EOF)
    {
        int i = 0;
        while (s[i] != it&&i<=46)
            i++;
        if (s[i])
        {
            cout<1];
        }
        else
        {
            cout<return 0;
}