401 - Palindromes

最新推荐文章于 2024-04-29 22:18:42 发布

原创最新推荐文章于 2024-04-29 22:18:42 发布 · 377 阅读

0 ·

CC 4.0 BY-SA版权

UVa 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一个C语言程序，该程序能够读取输入的字符串，并判断这些字符串是否为回文或者镜像字符串。通过对字符进行特定的映射转换，程序能够识别经过转换后的字符串是否与其反转形式相匹配。

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>

char * dic1 = "A   3  HIL JM O   2TUVWXY5";
char * dic2 = "1SE Z  8 ";

int main ( int argc, char * argv[] ) {

	char s[32];

	while( fgets ( s, 32, stdin ) != NULL ) {
		
		int len = strlen( s );
		if( len > 0 ) s[ --len ] = '\0';

		bool pal = true, mir = true;

		int i;

		for( i = 0; i < ceil( len / 2.0f ); i++ )
			if( s[ i ] != s[ len - 1 - i ] ) {
				pal = false;
				i = len;
			}

		for( i = 0; i < ceil( len / 2.0f ); i++ ) {
			char c1 = s[ i ];
			char c2 = s[ len - 1 - i ];

			if( c1 >= 'A' && c1 <= 'Z' ) c1 = dic1[ c1 - 'A' ];
			else if( c1 >= '1' && c1 <= '9' ) c1 = dic2[ c1 - '1' ];
			else { mir = false; i = len; }

			if( mir ) {
				if( c1 == ' ' || c1 != c2 ) {
					mir = false;
					i = len;
				}
			}

		}

		printf( "%s", s );
		if( !pal && !mir )
			printf( " -- is not a palindrome.\n\n" );
		else if( pal && !mir )
			printf( " -- is a regular palindrome.\n\n" );
		else if( !pal && mir )
			printf(" -- is a mirrored string.\n\n");
		else
			printf(" -- is a mirrored palindrome.\n\n");


	}
	return 0;
}