C 中数字的凯撒密码

Question

I want to make a Caesar cipher for numbers. (Add 3 to all digits)

Input: 52 Output: 85

Input: 954 Output: 287

Input: -10457 Output: -43780

I'll be very glad if someone helps me with this.

I tried this but when I input the number less than 5 digits it outputs 3 to beginning.

When I input 52 it outputs 33385. I also want to ask the user whether the program should accept a new number or just exit.

#include <stdio.h>
#include <stdlib.h>

int main()
{

    int number, operation;

    printf("Enter the number: ");
    scanf("%d", &number);

    printf("%d", ((number / 10000) + 3) % 10);
    printf("%d", (((number % 10000) / 1000) + 3) % 10);
    printf("%d", (((number % 1000) / 100) + 3) % 10);
    printf("%d", (((number % 100) / 10) + 3) % 10);
    printf("%d\n", ((number % 10) + 3) % 10);

    printf("press 1 to continue or 2 for exit.");
    scanf("%d", &operation);

    switch (operation) {
    case 1:
        printf("Enter the number: ");
        scanf("%d", &number);

        printf("%d", ((number / 10000) + 3) % 10);
        printf("%d", (((number % 10000) / 1000) + 3) % 10);
        printf("%d", (((number % 1000) / 100) + 3) % 10);
        printf("%d", (((number % 100) / 10) + 3) % 10);
        printf("%d\n", ((number % 10) + 3) % 10);

        break;

    case 2:

        break;
    }

    return 0;
}

Answer 1

When you enter 52, your number variable takes the value 00052, which will be transformed with your code into 33385. If you don't want the leading 3, you can either not print them, or retrieve the digit first and compute its associated Caesar digit only if the digit is not 0.

Another possibility to take into account any int number (whatever the number of digit ) is:

#include "stdio.h"
#include <stdlib.h>

int main()
{
    int number, abs_number;
    int caesar_number = 0;
    int digit_pos = 1;

    printf("Enter the number: ");
    scanf("%d", &number);

    abs_number = abs(number);

    while (abs_number != 0)
    {
        caesar_number += (((abs_number % 10) + 3) % 10) * digit_pos;
        abs_number /= 10;
        digit_pos *= 10;
    }

    if (number < 0)
    {
        caesar_number = -caesar_number;
    }
    printf("Result is: %d\n", caesar_number);

    return 0;
}

Answer 2

This solution works for all your test cases (positive and negative numbers), for numbers greater than 100000 and for zero as well:

#include "stdio.h"
#include "stdlib.h"

const int SHIFT = 3;

int main() {
    int number, abs_number, is_negative, operation, result, decimal_place;

    do {
        result = 0;
        decimal_place = 1;

        printf("Enter the number: ");
        scanf("%d", &number);

        // remove the sign before entering the cipher logic...
        abs_number = abs(number);
        // ...but do remember it so we can add it back later
        is_negative = number < 0;

        // from least to most significant decimal place
        do {
            // cipher the digit and accumulate it in the result
            result += (((abs_number % 10) + SHIFT) % 10) * decimal_place;
            // remove the consumed digit from the input variable
            abs_number /= 10;
            // move to the next decimal place
            decimal_place *= 10;
        } while (abs_number > 0);

        printf("Result is: %s%d\n", is_negative ? "-" : "", result);

        printf("Press 1 to continue or 2 for exit.");
        scanf("%d", &operation);

    } while (operation == 1);  // repeat if user requested another cipher

    return 0;
}

Some things to note in your original code:

• your code doesn't loop. A switch case won't help you repeat the operation. You need a while loop for that;
• it doesn't work for larger numbers: you would have to add an extra line of code to handle numbers greater than 100000... which leads to my next point:
• you have repeated code: your logic is duplicated in two different ways. The first one is the iteration body. Even if you replaced the switch statement with a while loop, your first iteration is unnecessarily unrolled outside of the loop. The second one is the way you handle each decimal place of the input number. Wouldn't it be better if you had a same code snippet to handle any of the decimal places? Duplicating code is considered bad practice. It makes the code bigger, it's harder to maintain and easy for duplicated code to get changed in only one place and not in the other. You always want to avoid duplicating code.

Answer 3

Unless the length of user input is not recorded, "Caesar's cipher for numbers" is not reversalable. With input like "777" , "7" , "000" , "0" , the output needs to be 4 distinctive answers in order to convert back.

To maintain something of a numeric internal representation, consider using "%n" to keep track of the length of user input.

static void print_digits(int a, int length, int offset) {
  if (length > 1) {
    print_digits(a/10, length-1);
  }
  putchar((a+offset)%10 + '0');
}

...

#define OFFSET 3

int number;
int n1, n2;
int offset = OFFSET % 10;
if (offset < 10) offset += 10; // Handle negative OFFSET

printf("Enter the number: ");
if (scanf(" %n%d%n", &n1, &number, &n2) == 1) {
  int length = n2 - n1;
  if (number < 0) {
    putchar('-');
    number = -number;
    length--;
  }
  print_digits(number, length, offset);
}

Above code has trouble with select input magnitudes above INT_MAX/10 or so. To handle large integers, consider moving to string processing.