uva401--字符串

字符串反转的题，做两个判断

1、判断是否为回文串

2、判断是否为反转串

两种情况分开做都很简单，合并在一起也比较简单

注意：奇偶情况略有差别

//#define LOCAL
#include <stdio.h>
#include <string.h>

#define MAXN 20 + 10

char map[] = {
	'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','1','2','3','4','5','6','7','8','9'
};

char reverse[] = {
	'A',' ',' ',' ','3',' ',' ','H','I','L',' ','J','M',' ','O',' ',' ',' ','2','T','U','V','W','X','Y','5','1','S','E',' ','Z',' ',' ','8',' '
};

char str[MAXN]; 

int find(char ch);
void palindromes();


int main()
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
#endif
	
	while(~scanf("%s", str))
	{
		// 算法主体
		palindromes(); 
		 
	}
	return 0;
}

void palindromes()
{
	bool flag1 = true;// 回文串标志位 
	bool flag2 = true;// 反转串标志位 
	for(int i = 0; i < strlen(str) / 2 + 1; i++)
	{
		// 判断是否为镜像串 
		int index = find(str[i]);
		char temp1 = reverse[index];
		char temp2 = str[strlen(str) - 1 - i]; 
		if(temp1 != temp2)
			flag1 = false;
			
		// 判断是否为回文串
		if(str[i] != str[strlen(str) - 1 - i] && i != (strlen(str) - 1 - i))
			flag2 = false; 	
	}
	
	if(flag1 && flag2)
		printf("%s -- is a mirrored palindrome.\n\n", str);
	else if(flag1) 
			printf("%s -- is a mirrored string.\n\n", str);
		else if(flag2)
				printf("%s -- is a regular palindrome.\n\n", str);
			else
				printf("%s -- is not a palindrome.\n\n", str);
}

int find(char ch)
{
	int len = strlen(map);
	for(int i = 0; i < len; i++)
		if(map[i] == ch)
			return i; 
	return -1;
}