spoj HS12HDPW - Hidden Password

最新推荐文章于 2024-08-19 17:50:15 发布

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最新推荐文章于 2024-08-19 17:50:15 发布

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本文介绍了一个关于从两个特定格式的字符串中提取隐藏密码的问题。通过位运算解析第一个字符串中的六字符元组，得到指示第二个字符串中字符位置的数值，从而揭示隐藏的密码。

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You are given two alphanumeric ASCII strings.An ancient manuscript says those strings contain a hidden password. Decode it!

The first string may be grouped into tuples of six characters each.For each such 6-tuple, taking from the i-th character (start counting from 0) the i-th bit of its ASCII code gives you a number (call it a), and likewise taking the (i+3 mod 6)-th bits gives you another number (call it b).

These two numbers tell you about the next two characters to be included in the password, namely the a-th and the b-th character from the second string (count starting from 0 as usual).

Input

First, you are given t (t<100) - the number of test cases.

Each of the test cases starts with one number n (n < 100) - the number of 6-tuples in the first string, followed by the two strings in separate lines (please have a look at the example to see the correct format).The second string is 64 characters long.

Successive test cases are separated by an empty line.

Output

For each of the test cases, output its hidden password in a separate line.

Example

Input:
2
2
qwe345 rf3Arg
XSBSRasdew9873465hkldsfsalndfvnfq489uqovkLKJHaeDaae555Sk5asdpASD

3
2S4J5K 111111 lrtb2A
isimgsow45ipfgisd56wfgngdfcdkgc7kKKKkuuJJgfstdygQdWORQADFSLKF2K8

Output:
keep
coding

Explanation
Let us have a look at the first 6-tuple: qwe345

char. ASCII code
  q   113 = 01110001B      
  w   119 = 01110111B
  e   101 = 01100101B
  3    51 = 00110011B
  4    52 = 00110100B
  5    53 = 00110101B

a(blue bits)  = 110111B = 55
b(green bits) = 101110B = 46

Scoring

By solving this problem you score 10 points.

题意：给出两行字符串，第一行字符串是由6个字符组成的字符串的元组，从每个元组中抽取第i个字符的第i位，构成一个数值，这个数值刚好是密文在第二个字符串的位置

思路：根据题意做就可以了，主要是位操作

代码如下：

<?php
	$file = STDIN;#fopen('./spoj.txt', 'r');
	$t = fgets($file);	
	$t = trim($t);
	while ($t--)
	{
		$n = trim(fgets($file));
		$s1 = trim(fgets($file));
		$s2 = trim(fgets($file));
		$arr = explode(' ', $s1);
		
		$ans = '';
		for ($i = 0; $i < $n; $i++)
		{
			$s = $arr[$i];
			$blue = 0; $green = 0;
			for ($j = 0; $j < 6; $j++)
			{
				$ascii = ord($s{$j});
				$bit = ($ascii >> $j) & 1;
				$blue |= $bit << $j;	
				$bit = ($ascii >> (($j + 3) % 6)) & 1;
				$green |= $bit << $j;							
			}
			$ans .= $s2{$blue};
			$ans .= $s2{$green};
		}	
		echo $ans.PHP_EOL;	
		if (0 != $t) fgets($file);
	}

	#fclose($file);