Answer / gganesh

#include<stdio.h>
#include<conio.h>
void main()
{
int n;
clrscr();
printf("Enter the number : ");
scanf("%d",&n);
n%2?printf("Odd"):printf("Even");
getch();
}

note: here i used conditional operator, not a relational
operators and control structures

Answer / vignesh1988i

THE LAST two answers posted by two folks are correct but the
declarations have been made wrong...... we cant make use of
1D array here , if so only 'e' or 'o' only will get
printed.... but that is not our aim... so correct
declaration is using a 2D array.....
char a[][6]={{"even"},{"odd"}};

and also it is not the must to make use of array of pointers
concept...........

thank u

Answer / sanju uthaiah

#include<stdio.h>

int main()
{
char result[2]={"Even","Odd"};
int n=40;
printf("%d is %s",n,result[n%2]);
return 0;
}

Answer / vishal jain

#include<stdio.h>
#include<conio.h>
int main()
{
int num;
printf("Enter any Number : \n");
scanf("%d",&num);
char *s[2]={"Even","Odd"};
printf("%s",s[num&1]);
return 0;
}

Answer / vishal jain

#include<stdio.h>
#include<conio.h>
int main()
{
int num,i;
printf("Enter any Number : \n");
scanf("%d",&num);
i=num&1;
if(i==1)
{
printf("ODD");
}
else
{
printf("EVEN");
}
return 0;
}

Answer / mohd parvez 09311349697

#include<stdio.h>
#include<conio.h>
void main()
{
int num;
printf("Enter a number:");
scanf("%d",&num);
num%2&&printf("Number is ODD")||printf("Number is EVEN");
getch();
}

Answer / rohit agarwal

#include<stdio.h>
#include<conio.h>
main()
{
int n;
char *p[]={"Even","odd"};
clrscr();
printf("Enter the number");
scanf("%d",&n);
n=n%2;
printf("The value is %s",p[n]);
getch();

Answer / om

void odd_even_check(int z)
{
(z&1)?printf("\nodd\n"):printf("\neven\n");
}

Answer / asad

int oddeven(int n)
{
if(n&1)
return 1; //odd

else
return 0; //even
}

Answer / satyanarayana

#include<stdio.h>
void main()
{
int p;
printf("number:");
scanf("%d",&p);
while(p%2)
{
printf("odd");
}
printf("even");
}

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