求1解密函数的对应加密函数

程序员文章站 2022-05-10 10:53:42

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求一解密函数的对应加密函数

本帖最后由 xuzuning 于 2013-02-21 10:07:35 编辑现在已知一解密函数（readkey）如下，求对应的加密函数


function authcode( $string, $operation = "DECODE", $key = "", $expiry = 0 )
{
				$ckey_length = 4;
				$key = md5( $key != "" ? $key : "a2264dxLupVMlkqR" );
				$keya = md5( substr( $key, 0, 16 ) );
				$keyb = md5( substr( $key, 16, 16 ) );
				$keyc = $ckey_length ? $operation == "DECODE" ? substr( $string, 0, $ckey_length ) : substr( md5( microtime( ) ), 0 - $ckey_length ) : "";
				$cryptkey = $keya.md5( $keya.$keyc );
				$key_length = strlen( $cryptkey );
				$string = $operation == "DECODE" ? base64_decode( substr( $string, $ckey_length ) ) : sprintf( "%010d", $expiry ? $expiry + time( ) : 0 ).substr( md5( $string.$keyb ), 0, 16 ).$string;
				$string_length = strlen( $string );
				$result = "";
				$box = range( 0, 255 );
				$rndkey = array( );
				$i = 0;
				for ( ;	$i 				{
								$rndkey[$i] = ord( $cryptkey[$i % $key_length] );
				}
				$j = $i = 0;
				for ( ;	$i 				{
								$j = ( $j + $box[$i] + $rndkey[$i] ) % 256;
								$tmp = $box[$i];
								$box[$i] = $box[$j];
								$box[$j] = $tmp;
				}
				$a = $j = $i = 0;
				for ( ;	$i 				{
								$a = ( $a + 1 ) % 256;
								$j = ( $j + $box[$a] ) % 256;
								$tmp = $box[$a];
								$box[$a] = $box[$j];
								$box[$j] = $tmp;
								$result .= chr( ord( $string[$i] ) ^ $box[( $box[$a] + $box[$j] ) % 256] );
				}
				if ( $operation == "DECODE" )
				{
								if ( ( substr( $result, 0, 10 ) == 0 || 0 								{
												return substr( $result, 26 );
								}
								return "";
				}
				return $keyc.str_replace( "=", "", base64_encode( $result ) );
}
function readkey( $keys )
{
		$Leskey = authcode( $keys, "DECODE", "a2264dxLupVMlkqR" );
		$Leskey = base64_decode( $Leskey );
		$firstStr = substr( $Leskey, 0, 1 );
		$Leskey = substr( $Leskey, 1 );
		$Leskey = strrev( substr( $Leskey, 0, $firstStr ) ).chr( ord( substr( $Leskey, $firstStr, 1 ) ) ^ $firstStr ).strrev( substr( $Leskey, $firstStr + 1 ) );
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