递归换位密码 C

Question

>我正在尝试输入一串字符并将它们递归地分成两半，直到字符长度仅为 2。然后取 2 的 char 字符串并将其与旁边的 char 字符串交换。我的代码如下：

#include <stdio.h>
#include <string.h>
void encrypt(char *str, int size);
int main(){
    char input[8192];
    int length;
    printf("INPUT A PHRASE: ");
    fgets(input, 8192, stdin);
    length = strlen(input) -1;
    printf("LENGTH: %dn", length);
    encrypt(input, length);
    printf("ENCRYPTION: %sn", input);  
    return 0;
}
void encrypt(char str[], int size){
    int i;
    int k = size / 2;
    //Encryption code here
}

一些示例输出：

Sample Input:
12345678
Sample Output:
34127856
Sample Input:
Test early and often!
Sample Output:
aeyrleT sttf!enn aod

我

不会要求任何人为我做作业。只是在寻找正确方向的推动，因为我很难弄清楚如何连续拆分字符串然后交换。

Answer 1

OP："我不会要求任何人为我做作业。只是在寻找朝着正确方向的推动

"

递归代码需要拆分字符串并注意长度，无论是奇数还是偶数。

void encrypt(char str[], int size){
  if (size <= 2) {
    // do encryption, TBD code
    retrun;
  }
  int i;
  int k = size / 2;  //  good step 1
  char *left = ____; // what is the address of the left half?  (easy)
  // call encrypt of the left half
  encrypt(left, ___); // what is the size of the left half? (easy)
  // Now do right
  char *right = ____; // what is the address of the R half? 
                      // (hint, it begins at the left half end.)
  // What is the R half length (hint: total - ???)
  // now call encrypt again with R half and length
  encrypt(right, ___);
}