poj 1590Palindromes 简单模拟

标签：poj   1590   palindromes   简单模拟   

//简单模拟吧，哎，一直感觉自己很水
//继续练吧

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define ceil(a,b) (((a)+(b)-1)/(b))
#define endl '\n'
#define gcd __gcd
#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))
#define popCount __builtin_popcountll
typedef long long ll;
using namespace std;
const int MOD = 1000000007;
const long double PI = acos(-1.L);

template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }
template<class T> inline T lowBit(const T& x) { return x&-x; }
template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }
template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }

char a[200];
char str[50];

void init(){
	memset(a,0,sizeof(a));
	a['A'] = 'A';
	a['E'] = '3';
	a['H'] = 'H';
	a['I'] = 'I';
	a['J'] = 'L';
	a['L'] = 'J';
	a['M'] = 'M';
	a['O'] = 'O';
	a['S'] = '2';
	a['T'] = 'T';
	a['U'] = 'U';
	a['V'] = 'V';
	a['W'] = 'W';
	a['X'] = 'X';
	a['Y'] = 'Y';
	a['Z'] = '5';
	a['1'] = '1';
	a['2'] = 'S';
	a['3'] = 'E';
	a['5'] = 'Z';
	a['8'] = '8';
	
}
int n;
bool ispalindrome(){
	int i=0, j=n-1;
	while(i<j){
		if (str[i]=='0')
			str[i] = 'O';
		if (str[j]=='0')
			str[j] = 'O';
		if (str[i]!=str[j])
			return false;
		i++;
		j--;
	}
	return true;
}

bool ismirrior(){
	int i=0,j=n-1;
	while(i<=j){
		if (str[i]=='0')
			str[i] = 'O';
		if (str[j]=='0')
			str[j] = 'O';
		if (a[str[i]]==0)
			return false;
		if (a[str[i]]!=str[j])
			return false;
		i++;
		j--;
	}
	return true;
}

int main() {
	init();
    freopen("G:\\Code\\1.txt","r",stdin);
	while(scanf("%s",str)!=EOF){
		n = strlen(str);
		bool flag1 = ispalindrome();
		bool flag2 = ismirrior();
		if (!flag1&&!flag2){
			printf("%s -- is not a palindrome.",str);
		}else if (!flag1&&flag2){
			printf("%s -- is a mirrored string.",str);
		}else if (flag1&&!flag2){
			printf("%s -- is a regular palindrome.",str);
		}else {
			printf("%s -- is a mirrored palindrome.",str);
		}
		puts("");
		puts("");
	}
	return 0;
}

poj 1590Palindromes 简单模拟

标签：poj   1590   palindromes   简单模拟   

原文地址：http://blog.csdn.net/timelimite/article/details/44979245