Imprimer le chemin le plus court pour imprimer la chaîne à l'écran dans le programme C.

Given a string, the program must display the shortest path which will print the string over the screen using that shortest path.

Like screen will store alphabets in the format

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

Example

的中文翻译为：

示例

Input: HUP
Output : Move Down
Move Down
Move Down
destination reached
Move Left
Move Left
Move Down
Move Down
Move Down
destination reached
Move Up
destination reached

这里使用的方法是将字符存储在n x n矩阵中，并执行以下操作 −

If row difference is negative then move up
If row difference is positive then move down
If column difference is negative then go left
If column difference is positive then we go right

算法

START
Step 1 -> Declare Function void printpath(char str[])
   Declare variable int i = 0 and cx=0 and cy=0
   Loop While str[i] != '\0'
      Declare variable as int n1 = (str[i] - 'A') / 5
      Declare variable as int n2 = (str[i] - 'B' + 1) % 5
      Loop while cx > n1
         Print move up
         cx—
      End
      Loop while cy > n2
         Print Move Left
         Cy—
      End
      Loop while cx < n1
         Print move down
         Cx++
      End
      Loop while cy < n2
         Print move down
         Cy++
      End
      Print destination reached
      I++
Step 2 -> in main()
   Declare char str[] = {"HUP"}
   Call printpath(str)
STOP

Example

的中文翻译为：

示例

#include <stdio.h>
void printpath(char str[]){
   int i = 0;
   // start from character &#39;A&#39; present at position (0, 0)
   int cx = 0, cy = 0;
   while (str[i] != &#39;\0&#39;){
      // find cordinates of next character
      int n1 = (str[i] - &#39;A&#39;) / 5;
      int n2 = (str[i] - &#39;B&#39; + 1) % 5;
      // Move Up if destination is above
      while (cx > n1){
         printf("Move Up</p><p>");
         cx--;
      }
      // Move Left if destination is to the left
      while (cy > n2){
         printf("Move Left</p><p>");
         cy--;
      }
      // Move down if destination is below
      while (cx < n1){
         printf("Move Down</p><p>");
            cx++;
      }
      // Move Right if destination is to the right
      while (cy < n2){
         printf("Move Down</p><p>");
         cy++;
      }
      // At this point, destination is reached
      printf("destination reached</p><p>");
      i++;
   }
}
int main(int argc, char const *argv[]){
   char str[] = {"HUP"};
   printpath(str);
   return 0;
}

输出

如果我们运行上面的程序，它将生成以下输出−

Move Down
Move Down
Move Down
destination reached
Move Left
Move Left
Move Down
Move Down
Move Down
destination reached
Move Up
destination reached

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