The question says to write a program asking the user to enter 2 digit number, then prints the English word for it. Suppose you enter \'41\' the printf function prints out \'
@user2409559 , @Dinever; First,no need of % in ur code as u are taking i/p as first and second digits. Second your code is incomplete. Enter a number b/w 10 to 19 and see the o/p.
I am posting my code below.......take a look.
#include
#include
int main()
{
int num,n,r;
printf("Enter a two-digit number: ");
scanf("%d",&num);
n = num/10;
r = num%10;
switch(n)
{
case 1: switch(r)
{
case 0: printf("Ten");
break;
case 1: printf("Eleven");
break;
case 2: printf("Twelve");
break;
case 3: printf("Thirteen");
break;
case 4: printf("Fourteen");
break;
case 5: printf("Fifteen");
break;
case 6: printf("Sixteen");
break;
case 7: printf("Seventeen");
break;
case 8: printf("Eighteen");
break;
case 9: printf("Nineteen");
}
break;
case 2: printf("Twenty-");
break;
case 3: printf("Thirty-");
break;
case 4: printf("Fourty-");
break;
case 5: printf("Fifty-");
break;
case 6: printf("Sixty-");
break;
case 7: printf("Seventy-");
break;
case 8: printf("Eighty-");
break;
case 9: printf("Ninety-");
break;
}
if(n != 1)
{
switch(r)
{
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: ;
}
}
getch();
}