Question

我正在尝试接收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char字符串2并将其与旁边的字符串交换。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: %d\n", length);

    encrypt(input, length);
    printf("ENCRYPTION: %s\n", 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

我不是要求任何人为我做任务。只是想在正确的方向上轻推一下，因为我正在艰难地绕着如何连续分开琴弦然后交换。

Answer 1

OP: "I'm not asking anyone to do my assignments for me. Just looking for a nudge in the right direction"

The recursive code needs to split the string and be careful about the length, be it odd or even.

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, ___);

}

递归转置密码C.

1 个答案: