UVA - 401 Palindromes

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UVA - 401 Palindromes

题目大意：给定一个字符串判断是否回文是什么样打回文

解题思路：。。强行扫

#include <iostream>#include <cstdio>using namespace std;int main () {    string a,c;    bool f;    int t = 1;    while (cin>>a ) {        c = a;        f = 1;        int p = 0;        for (int i = 0; i < a.length(); i++) {             if ( (a[i] == 'A' || a[i] == 'H' || a[i] == 'I' || a[i] == 'M' || a[i] == 'O' || a[i] == '0' || a[i] == 'T' || a[i] == 'U' || a[i] == 'V' || a[i] == 'W' || a[i] == 'X' || a[i] == 'Y' || a[i] == '8') && p != 2 && p != 3) {                p = 1;                if ( a[i] == '0')                    a[i] == 'O';            }            else if ( a[i] == 'E'  && a[a.length()- 1 - i] == '3'|| a[i] == '3'  && a[a.length()- 1 - i] == 'E' || a[i] == 'J'  && a[a.length()- 1 - i] == 'L' || a[i] == 'L'  && a[a.length()- 1 - i] == 'J'|| a[i] == 'S'  && a[a.length()- 1 - i] == '2'|| a[i] == '2'&& a[a.length()- 1 - i] == 'S'|| a[i] == 'Z' && a[a.length()- 1 - i] == '5' || a[i] == '5' && a[a.length()- 1 - i] == 'Z' || a[i] == '1' && a[a.length() - i -1] == '1') {                p = 2;                if ( a[i] == '3') a[i] = 'E';                if ( a[i] == 'L') a[i] = 'J';                if ( a[i] == '2') a[i] = 'S';                if ( a[i] == '5') a[i] = 'Z';             }            else if ( p != 2) p = 3;        }        for (int i = 0; i < a.length();i++)        if ( a[i] != a[a.length() - 1 - i]) p = 0;      //  if (f) {            if (p == 1 )cout<<c<<" -- is a mirrored palindrome."<<endl;            else if (p == 2)cout<<c<<" -- is a mirrored string."<<endl;                else if ( p == 3 ) cout<<c<<" -- is a regular palindrome."<<endl;//        } else cout<<c<<" -- is not a palindrome."<<endl;        cout<<endl;    }}

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