我正在编写一个代码来显示C语言中任何正整数的数字.例如,整数345将显示为三个四五.
我编写的代码适用于所有数字大于0的整数.但是某些整数(如10,304,0等)无法正确显示.为了挑战,不允许使用递归和数组.只允许循环和if构造.有什么建议?
#include<stdio.h>
int main(void)
{
int num, num_copy, accum = 1;
printf("Enter an integer ");
scanf("%i",&num);
num_copy = num;
while (num > 0){
while (num > 9){
num = num / 10;
accum *= 10;
}
switch (num){
case (1):
printf("one ");
break;
case (2):
printf("two ");
break;
case (3):
printf("three ");
break;
case (4):
printf("four ");
break;
case (5):
printf("five ");
break;
case (6):
printf("six ");
break;
case (7):
printf("seven ");
break;
case (8):
printf("eight ");
break;
case (9):
printf("nine ");
break;
}
num_copy = num_copy - (num*accum);
num = num_copy;
accum = 1;
}
return 0;
}
Run Code Online (Sandbox Code Playgroud)
哦!这听起来很有趣!没有数组也没有递归,对吗?
因此,由于我们不能使用递归,我们需要向后构建数字:
#include<stdio.h>
int main(void)
{
int num, backwards = 0, digit, backupzeros = 0;
printf("Enter an integer ");
scanf("%i",&num); // get the number
while(num > 0){
digit = num % 10; // pry off the last digit
num /= 10; // take off the digit
if((backwards == 0) && (digit == 0)) // If it's a number that ends in 0
backupzeros++; // save it, we'll use that later
backwards = (backwards * 10) + digit; // glue it on backwards
}
// Now we have the number reversed. Next we need to print the digits
while (backwards > 0){
digit = backwards % 10;
backwards /= 10;
switch (digit){
case 1:
printf("one ");
break;
case 2:
printf("two ");
break;
case 3:
printf("three ");
break;
case 4:
printf("four ");
break;
case 5:
printf("five ");
break;
case 6:
printf("six ");
break;
case 7:
printf("seven ");
break;
case 8:
printf("eight ");
break;
case 9:
printf("nine ");
break;
default:
printf("zero ");
break;
}
}
while(backupzeros > 0) {
printf("zero ");
backupzeros--;
}
return 0;
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
2727 次 |
| 最近记录: |