Question

Please write code in C, thank you.

Write a program that reads a list of integers, and outputs whether the list contains all even numbers, odd numbers, or neither. The input begins with an integer indicating the number of integers that follow. Assume that the list will always contain less than 20 integers.

Ex: If the input is:

5 2 4 6 8 10

the output is:

all even

Ex: If the input is:

5 1 3 5 7 9

the output is:

all odd

Ex: If the input is:

5 1 2 3 4 5

the output is:

not even or odd

Your program must define and call the following two functions. IsArrayEven returns true if all integers in the array are even and false otherwise. IsArrayOdd returns true if all integers in the array are odd and false otherwise.
bool IsArrayEven(int inputVals[], int numVals)
bool IsArrayOdd(int inputVals[], int numVals)

Answer 1

Code:

#include<stdio.h>
#include<stdbool.h>                   //for boolean
bool IsArrayOdd(int array[],int n)
{
   for(int i=0;i<n;i++)
   {
       if(array[i]%2==0)           //if any even number encounter then return false
           return false;
   }
   return true;                   //if no even number then return true
}
bool IsArrayEven(int array[],int n)
{
   for(int i=0;i<n;i++)
   {
       if(array[i]%2!=0)           //if any odd number encounter then return false
           return false;
   }
   return true;                   //if no odd number then return true
}
int main()
{
   int n;
   scanf("%d",&n);                   //size of array
   int array[n];
   int i,j;
   for(i=0;i<n;i++)
   {
       scanf("%d",&array[i]);       //array input
   }
   i=IsArrayOdd(array,n);           //calling function
   j=IsArrayEven(array,n);           //calling function
   if(j)                           //if j is true
       printf("all even\n");
   else if(i)                       //if i is true
       printf("all odd\n");
   else                           //if i and j are false
       printf("not even or odd\n");
}