我正在尝试创建一个简单的加密程序，允许用户输入一个5个字母的单词作为键。然后，程序使用该密钥的每个字符来加密他们输入的用户消息的字母。这有点难以解释，但一个例子将有助于使其更清晰。
所以，说你希望你的关键词是“柠檬”。 'l'是字母表中的第12个字母，但我们从0开始计数，所以'l'是我们称之为字母数组的第11个字母。'l' = 11, 'e' would = 4, 'm' = 12, 'o' = 14, 'n' = 13。因此，如果您想加密消息“黎明时分的攻击”，每个角色都会拥有自己对应的密钥。然后，它会将与密钥相对应的数字添加到消息字母中，然后对其进行加密。示例如下。
plain text: attack at dawn
key: lemonl em onle
encrypted: lxfopv ef rnhr
查看代码和注释也会有所帮助。我的问题是，应该实际加密消息的for循环不能正常工作。我看不出代码有什么问题，但是输出不正确所以它一定是逻辑错误吗？
#include <iostream>
#include <string>
using namespace std;
int main() {
string key;
int keySize;
char keyChar[5];
int keyInteger[5];
bool keyLoop = true;
//makes sure key is 5 characters longs.
while (keyLoop) {
cout << "enter in the 5 letter key: ";
cin >> key;
keySize = key.size();
if (keySize < 5 || keySize > 5) {
cout << "Invalid key." << endl;
}
else {
//obtains each individual character and places it into the array keyChar.
for (int i = 0; i < 5; i++) {
keyChar[i] = key[i];
}
keyLoop = false;
}
}
char alphabet[52]{ 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',
'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z' };
//finds the place of the element in keyChar and stores it as an int in the array keyInteger. This is setting up the key for encryption.
for (int x = 0; x < 5; x++) {
for (int i = 0; i < 26; i++) {
if (keyChar[x] == alphabet[i]) {
keyInteger[x] = i;
}
}
}
//This just displays the values of keyInt. This code was included to make sure the program was storing the right values.
for (int i = 0; i < 5; i++) {
cout << keyInteger[i] << endl;
}
string secretMessage;
cin.ignore();
//user input their secret message
cout << "Enter your secret message: ";
getline(cin, secretMessage);
//gets size of message and creates variable to modify when we're ready to encrypt the message.
int secretMessageSize = secretMessage.size();
cout << secretMessageSize << endl;
string newMessage = secretMessage;
int temp;
// This first loop goes through each character of the message.
for (static int y = 0; y < secretMessageSize; y++) {
// This loop keeps track of which keyInteger will be used for the encryption formula.
for (int a = 0; a < 5; a++) {
// The if statement makes sure that the keyInteger isn't used until there is actually a character to be modified.
if (secretMessage[y] == ' ') {
//Just outputting the new message right now to make sure the program is running.
cout << " ";
a--;
}
else {
//this loop goes runs until the character of secretMessage[y] is the same as the character at alphabet[i].
//then it assigns temp to the value of i. This will be used in the encryption formula.
for (int i = 0; i < 26; i++) {
if (secretMessage[y] == alphabet[i]) {
temp = i;
break;
}
//this is all for testing purposes. LETTER should have the same letter in a row.
cout << "LETTER " << secretMessage[y] << " " << alphabet[i] << endl;
//number should have the what number character the loop is on of the secretMessage. The number across should be the where that character was found in the alphabet array.
cout << "NUMBER " << y << " " << i << endl;
//Key should only have one integer next to it. This should be whichever instance the loop that has a in it is at.
cout << "KEY " << keyInteger[a] << endl;
//This should give the actual character that should be replacing the character in secretMessage at whatever iteration it is at.
//The encryption formula takes the integer that is used to address the character in the alphabet array. So if it were an 'a', then that integer would == 0;
//Then it adds the keyInteger at whatever iteration the loop is at. If the keyword were lemon, and it was in its first iteration, a would = 0, and keyInteger[0] would equal 11.
//The sum of these two values will give you the new character from the alphabet array. So alphabet[0+11] would = l. <-- this is a lower case L.
cout << "CRYPT " << alphabet[keyInteger[a] + i] << endl;
}
}
//this adds to y so that it keeps track of each character in the message, but we are also able to keep track of which keyInteger to use.
y++;
}
}
return 0;
}
答案 0 :(得分：-2)
下面是用C ++编写的代码，它完全符合您的要求。如果有什么不起作用，你可以在评论中告诉我。
我使用 find（）函数以最小化代码长度并提高可读性。我主要修改了加密消息的 for（）循环（如你所提到的那样）。
编辑 - 加密时不要寻找空格。除了字母之外的任何其他内容都应该打印在加密消息中。我已使用 return（-1）执行相同的操作，并在加密循环中的 if condition 中进行了检查。
char alphabet[26]{ 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
#include <iostream>
#include <string>
using namespace std;
int find(char character)
{
for (int i = 0; i < 26; i++)
{
if (character == alphabet[i])
{
return (i);
}
}
return (-1);
}
int main()
{
string key;
int keySize;
char keyChar[5];
bool keyLoop = true;
int keyInteger[5];
while (keyLoop)
{
cout << "enter in the 5 letter key: ";
cin >> key;
keySize = key.size();
if (keySize < 5 || keySize > 5)
{
cout << "Invalid key." << endl;
}
else
{
for (int i = 0; i < 5; i++)
{
keyChar[i] = key[i];
}
keyLoop = false;
}
}
for (int x = 0; x < 5; x++)
{
keyInteger[x] = find(keyChar[x]);
}
string secretMessage;
cin.ignore();
cout << "Enter your secret message: ";
getline(cin, secretMessage);
int secretMessageSize = secretMessage.size();
string newMessage = secretMessage;
int temp;
int x = 0;
for (int i = 0; i < secretMessageSize; i++)
{
if (x == 5)
x = 0;
if (find(secretMessage[i]) == -1)
newMessage[i] = secretMessage[i];
else
{
temp = (find(secretMessage[i]) + keyInteger[x])%26;
newMessage[i] = alphabet[temp];
x++;
}
}
for (int i = 0; i < secretMessageSize; i++)
{
cout << newMessage[i];
}
cout << endl;
}