Project Euler：Problem 59 XOR decryption

Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk
(*) = 42, and lowercase k = 107.
A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function
is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.
For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption
key in different locations, and without both "halves", it is impossible to decrypt the message.
Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the
key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.
Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher.txt (right
click and 'Save Link/Target As...'), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.

看到题目想到了维吉尼亚密码，秘钥长度为3，显然每隔三位的字符都是用同一个字母加密的，并且这同一字符加密的字符按理应该是满足英文字母使用频率分布表的。

事与愿违！这统计出来的出现频率最高的字符与e的ASCII异或得到的东西很奇怪啊！可能是字符数太小的缘故。

还好秘钥长度为3，来个三层循环便历一下就好了。如果在加密过程中出现了除字母和标点符号以外的字符的话，说明这个秘钥不正确；如果像这种常见的字符比如,、.、the、that、is、to、it、in等在译出的明文中没有出现的话说明这个秘钥也是不正确的。

#include <iostream>
#include <fstream>
#include <string>
#include <vector>
using namespace std;

int main()
{
	ifstream input;
	input.open("cipher.txt");
	string src;
	input >> src;
	vector<char> asc;
	int a = 0;
	for (int i = 0; i < src.length(); i++)
	{
		if (src[i] == ',')
		{
			asc.push_back(a);
			a = 0;
		}
		else
		{
			int b = src[i] - '0';
			a = a * 10 + b;
		}
	}
	asc.push_back(a);
	vector<char> msg = asc;
	char key[3];
	bool the = false, and = false, to = false;
	int sum = 0;
	for (int x = 'a'; x <= 'z'; x++)
	{
		for (int y = 'a'; y <= 'z'; y++)
		{
			for (int z = 'a'; z <= 'z'; z++)
			{
				key[0] = x;
				key[1] = y;
				key[2] =  z;

				for (int i = 0; i < asc.size(); i++)
				{
					msg[i] = asc[i] ^ key[i % 3];
					if (i >= 2)
					{
						the = (msg[i - 2] == 't'&&msg[i - 1] == 'h'&&msg[i] == 'e') || the;
						and = (msg[i - 2] == 'a'&&msg[i - 1] == 'n'&&msg[i] == 'd') || and;
						to = (msg[i - 1] == 't'&&msg[i] == 'o') || to;
					}
					sum += msg[i];
				}
				if (the&&and&&to)
				{
					cout << sum << endl;
					for (int j = 0; j < msg.size(); j++)
						cout << msg[j];
					cout << endl;
				}
				the = false;
				and = false;
				to = false;
				sum = 0;
			}
		}
	}

	system("pause");
	return 0;
}

总共会输出四个结果，然后里面只有一个结果是能读懂的明文。