使用 ! 加密字符串的程序和 @
给定一个字符串，任务是使用!加密这个字符串和@符号，或者。在加密消息时，加密格式必须按照字母顺序重复符号与字母位置一样多的次数。
例子：
Input: string = "Ab"
Output: !@@
Explanation:
Position of 'A' in alphabetical order is 1
and in String is odd position
so encrypted message will have 1 '!'
Position of 'b' in alphabetical order is 2
and in String is even position
so encrypted message will have 2 '@'
Therefore, the output "!@@"
Input: string = "CDE"
Output: !!!@@@@!!!!!方法：这是一种非常基本和简单的加密技术，可以按如下方式完成：
- 从String中一一获取字符
- 对于每个字符，获取该字符的 ASCII 值与“A”（如果字符是大写字母）或“a”（如果字母是小写字母）之间的差异。这将是重复加密字符的次数。
- 对于字符串的第 i 个字符，如果 i 为奇数，则加密字符为 '!'如果我是偶数，加密字符将是'@'。
下面是上述代码的实现：
C++
// C++ program to Encrypt the String
// using ! and @
#include
using namespace std;
// Function to encrypt the string
void encrypt(char input[100])
{
// evenPos is for storing encrypting
// char at evenPosition
// oddPos is for storing encrypting
// char at oddPosition
char evenPos = '@', oddPos = '!';
int repeat, ascii;
for (int i = 0; i <= strlen(input); i++)
{
// Get the number of times the character
// is to be repeated
ascii = input[i];
repeat = ascii >= 97 ? ascii - 96 : ascii - 64;
for (int j = 0; j < repeat; j++)
{
// if i is odd, print '!'
// else print '@'
if (i % 2 == 0)
cout << oddPos;
else
cout << evenPos;
}
}
}
// Driver code
int main()
{
char input[100] = { 'A', 'b', 'C', 'd' };
// Encrypt the String
encrypt(input);
return 0;
}
// This code is contributed by
// shubhamsingh10 C
// C program to Encrypt the String
// using ! and @
#include
#include
// Function to encrypt the string
void encrypt(char input[100])
{
// evenPos is for storing encrypting
// char at evenPosition
// oddPos is for storing encrypting
// char at oddPosition
char evenPos = '@', oddPos = '!';
int repeat, ascii;
for (int i = 0; i <= strlen(input); i++) {
// Get the number of times the character
// is to be repeated
ascii = input[i];
repeat = ascii >= 97 ? ascii - 96 : ascii - 64;
for (int j = 0; j < repeat; j++) {
// if i is odd, print '!'
// else print '@'
if (i % 2 == 0)
printf("%c", oddPos);
else
printf("%c", evenPos);
}
}
}
// Driver code
void main()
{
char input[100] = { 'A', 'b', 'C', 'd' };
// Encrypt the String
encrypt(input);
} Java
// Java program to Encrypt the String
// using ! and @
class GFG
{
// Function to encrypt the string
static void encrypt(char input[])
{
// evenPos is for storing encrypting
// char at evenPosition
// oddPos is for storing encrypting
// char at oddPosition
char evenPos = '@', oddPos = '!';
int repeat, ascii;
for (int i = 0; i < input.length; i++)
{
// Get the number of times the character
// is to be repeated
ascii = input[i];
repeat = ascii >= 97 ?
ascii - 96 : ascii - 64;
for (int j = 0; j < repeat; j++)
{
// if i is odd, print '!'
// else print '@'
if (i % 2 == 0)
System.out.printf("%c", oddPos);
else
System.out.printf("%c", evenPos);
}
}
}
// Driver code
public static void main(String[] args)
{
char input[] = { 'A', 'b', 'C', 'd' };
// Encrypt the String
encrypt(input);
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 program to Encrypt the String
# using ! and @
# Function to encrypt the string
def encrypt(input_arr) :
# evenPos is for storing encrypting
# char at evenPosition
# oddPos is for storing encrypting
# char at oddPosition
evenPos = '@'; oddPos = '!';
for i in range(len(input_arr)) :
# Get the number of times the character
# is to be repeated
ascii = ord(input_arr[i]);
repeat = (ascii - 96 ) if ascii >= 97 \
else (ascii - 64);
for j in range(repeat) :
# if i is odd, print '!'
# else print '@'
if (i % 2 == 0) :
print(oddPos, end = "");
else :
print(evenPos, end = "");
# Driver code
if __name__ == "__main__" :
input_arr = [ 'A', 'b', 'C', 'd' ];
# Encrypt the String
encrypt(input_arr);
# This code is contributed by AnkitRai01C#
// C# program to Encrypt the String
// using ! and @
using System;
using System.Collections.Generic;
class GFG
{
// Function to encrypt the string
static void encrypt(char []input)
{
// evenPos is for storing encrypting
// char at evenPosition
// oddPos is for storing encrypting
// char at oddPosition
char evenPos = '@', oddPos = '!';
int repeat, ascii;
for (int i = 0; i < input.Length; i++)
{
// Get the number of times the character
// is to be repeated
ascii = input[i];
repeat = ascii >= 97 ?
ascii - 96 : ascii - 64;
for (int j = 0; j < repeat; j++)
{
// if i is odd, print '!'
// else print '@'
if (i % 2 == 0)
Console.Write("{0}", oddPos);
else
Console.Write("{0}", evenPos);
}
}
}
// Driver code
public static void Main(String[] args)
{
char []input = { 'A', 'b', 'C', 'd' };
// Encrypt the String
encrypt(input);
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
!@@!!!@@@@