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法律研究打印屏幕上字符串的最短路径的 C 程序。
给定一个字符串，该程序必须显示在屏幕上打印该字符串的最短路径，使用该最短路径。
屏幕将以这种格式存储字母
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
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示例
#include <stdio.h>
void printpath(char str[]){
int i = 0;
// start from character 'A' present at position (0, 0)
int cx = 0, cy = 0;
while (str[i] != '\0'){
// find cordinates of next character
int n1 = (str[i] - 'A') / 5;
int n2 = (str[i] - 'B' + 1) % 5;
// Move Up if destination is above
while (cx > n1){
printf("Move Up
");
cx--;
}
// Move Left if destination is to the left
while (cy > n2){
printf("Move Left
");
cy--;
}
// Move down if destination is below
while (cx < n1){
printf("Move Down
");
cx++;
}
// Move Right if destination is to the right
while (cy < n2){
printf("Move Down
");
cy++;
}
// At this point, destination is reached
printf("destination reached
");
i++;
}
}
int main(int argc, char const *argv[]){
char str[] = {"HUP"};
printpath(str);
return 0;
}输出
如果我们运行上述程序，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
更新于:2019-08-22
233 次观看
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