Consider the code below a) Modify the code to insert another elif case so that i
ID: 1765983 • Letter: C
Question
Consider the code below
a) Modify the code to insert another elif case so that if a non-integer is entered, the output text indicates that the input value is not an integer. Show the new code
b) Modify the code to use a try-except block so that the program will try to input a value into nn and perform the if-else structure. If a non-number is entered, output the text, “Input is not a number”. Show the new code
# DiVBy3.py-Krista Hill-Feb. 4, 2018 # Example collection of functions nn = input ("Integer: ") print nn, "is non-zero and is divisible by 3'" print nn, "is zero" print nn,"is neither zero or divisible by 3" else:Explanation / Answer
a answer:-
# collection of functions
nn = input("enter a number: ")
temp == nn
qq = nn - temp
if qq != 0
print nn, "not an integer"
elif nn!=0 and nn%3==0
print nn, "is a integer and divisible by 3"
elif nn==0
print nn, " is zero"
else:
print nn, " is neither zero or divisble by 3"
b answer:-
#include<stdio.h>
#include<stdlib.h>
int main(void){
int a;
float b,c;
printf("Enter the number ");
scanf("%f",&b);
a=b;
printf("a=%d ",a);
printf("b=%d ",b);
c=b-a;
b=b-a;
printf("a=%d ",a);
printf("b=%f ",b);
printf("c=%f ",c);
if (c==0){
printf("entered number is an interger ");
if(a==0)
printf("it is zero ");
else if(a!=0 && a%3==0)
printf("it is divisible by 3"):
else
printf("entered is non zero but not divisible by 3");
}
else
printf(" it is not an integer");
return o;
}