uva_850 - Crypt Kicker II

最新推荐文章于 2021-08-03 23:56:19 发布

原创最新推荐文章于 2021-08-03 23:56:19 发布 · 755 阅读

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#破解 #算法 #c

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本文介绍了一个简单的密文破解算法，通过枚举方法来寻找特定明文'the quick brown fox jumps over the lazy dog'在密文中对应的关系，进而解码整个密文。文章包含完整的C语言代码实现。

题意求密文破解有一段明文是一定会出现的 the quick brown fox jumps over the lazy dog // 有狗狗呦

能破解输出破解后的明文！否则输出 No solution.

算法:简单枚举

连接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=791

#include <stdio.h>
#include <string.h>
#define MAX 105
int main ( )
{
	int t;
	scanf ( "%d\n", &t );
	while ( t-- )
	{
		char str[MAX][MAX], visit[50];
		strcpy ( visit, "the quick brown fox jumps over the lazy dog" );
		int sum = 0, len = strlen ( visit ), hash[27] = { 0 };
		bool flag = false;
		while ( gets ( str[sum] ) )
			if ( strlen ( str[sum] ) < 1 )
				break;
			else
				sum++;
		for ( int i = 0; i < sum; i++ )
		{
			int n = strlen ( str[i] );
			memset ( hash, 0, sizeof hash );
			if ( n != len )
				continue;
			for ( int j = 0; j < len; j++ )
			{
				if ( str[i][j] == ' ' && visit[j] == ' ' )
					continue ;
				if ( 'a' <= str[i][j] && str[i][j] <= 'z' && 'a' <= visit[j] && visit[j] <= 'z' )
					if ( hash[str[i][j]-'a'+1] == 0 || hash[str[i][j]-'a'+1] == visit[j] - 'a' + 1 )
						hash[str[i][j]-'a'+1] = visit[j] - 'a' + 1;
					else
						break;
				else
				{
					if ( str[i][j] != visit[j] )
						break;
				}
			}
			flag = true;
			for ( int k = 1; k < 26; k++ )
				if ( !hash[k] )
					flag = false;
			if ( flag )
				break;
		}
		if ( flag )
		{
			for ( int i = 0; i < sum; i++ )
			{
				len = strlen ( str[i] );
				for ( int j = 0; j < len; j++ )
					if ( str[i][j] == ' ' )
						printf ( " " );
					else
						printf ( "%c", hash[str[i][j]-'a'+1] - 1 + 'a' );
				puts ( "" );
			}
		}
		else
			puts ( "No solution." );
		if ( t )
			puts ( "" );
	}
	return 0;
}