Answer / lakshmi

#include<studio.h>
#include<conio.h>
main()
{
int a[5],count_even=0,count_odd=0,i;

for(i=0;i<5;i++)

scanf("%d",&a[i]);
/* display the number of odd and even intergers */
for(i=0;i<5;i++)
{
if((a[i]%2 ==0))
count_even++;
if((a[i]%2==1))
count_odd++;
}
printf("%d %d",count_even,count_odd);
getch();
}

Answer / rahul khare

#include<stdio.h>
#include<conio.h>
void main()
{
int a[10],even=0,odd=0,i;
for(i=0;i<9;i++)
{
if(a[i]%2==0)
{
printf("%d\t",a[i]);
even++;
}
else
{
printf("%d\t",a[i]);
odd++;
}
}
printf("Total no of Even found is=%d",even);
printf("Total no of Odd found is=%d",odd);
getch();
}

Answer / shuvendu sekhar sahu

#include<stdio.h>
#include<conio.h>
main()
{
int x;
clrscr();
printf("Enter the value");
scanf("%d",& x);
if(x%2==0)
printf("x is even");
else
printf("x is odd");
getch();
}

Answer / anup majhi

#include<stdio.h>
#include<conio.h>
#define size 100
void check(int[],int);
int i;
main(){
int a[size],m
printf("Enter the array length ");
scanf("%d",&m);
printf("Supply elements to the array ");
for(i=0;i<m;i++)
scanf("%d",&a[i]);
check(a,m);
}
void check(int a[],int b){
int even=0,odd=0;
for(i=0;i<b;i++){
if(a[i]%2==0)
even++;
else
odd++;
}
printf("Number of even no. in the array =%d ",even);
printf("\nNumber of odd no. in the array =%d ",odd);
}


out put:
Enter the array length 5
Supply array elements 1 2 3 4 5
Number of even no. in the array =2
Number of odd no. in the array =3

Answer / ravestar

#include<stdio.h>
#include<iostream.h>
#include<conio.h>
main()
{clrscr();
int a[5],count_even=0,count_odd=0,i;

for(i=0;i<5;i++)

scanf("%d",&a[i]);
for(i=0;i<5;i++)
{
if((a[i]%2 ==0))
count_even++;
if((a[i]%2==1))
count_odd++;
}
cout<<"Even:"<<count_even<<"\n"<<"Odd:"<<count_odd;
getch();
}

Answer / ankit kr. sharma

#include<stdio.h>
#include<conio.h>
void main()
{

int a[10],i,neg=0,even=0,odd=0;
printf("Enter the elements for the array:\n");
for(i=0;i<10;i++)
{
scanf("%d",&a[i]);
}
printf("The array is:");
for(i=0;i<10;i++)
{
printf("%d\t", a[i]);
}
for(i=0;i<10;i++)
{
if (a[i]<0)
neg++;
else if(a[i]%2==0)
even++;
else
odd++;
}

printf("\nthe odd no.s in the array are: %d\t",odd);
printf("\nthe negative no.s in the array are: %d\t",neg);
printf("\nthe even no.s in the array are: %d\t",even);
}

Answer / ankit kumar sharma

#include<stdio.h>
#include<conio.h>
void main()
{
int a[5],count_even=0,count_odd=0,i;

printf("Enter the value in Array\n");
for(i=0;i<5;i++)
scanf("%d",&a[i]);
/* Calculating number of odd and even intergers */
for(i=0;i<5;i++)
{
if((a[i]%2 ==0))
count_even++;
if((a[i]%2==1))
count_odd++;
}
/* display the number of odd and even intergers */
printf(" Total number of even integer are %d\n ",count_even);
printf("Total number of odd integer are %d", count_odd);
getch();
}

Answer / merlin

#include<studio.h>
#include<conio.h>
void main()
{
int a[20],even=0,odd=0,i,n;
printf("Enter the size of the array");
scanf("%d",&n);
printf("Enter the elements");
for(i=0;i<n;i++)
{

scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{

if((a[i]%2 ==0))
even++;
else
odd++;
}
printf("%d %d",even,odd);
getch();
}

Answer / ks djd

#include<stdio.h>
void main()
{
int a[10],i,b;
for(i=0;i<10;i++)
{scanf("%d",&a[i]);
}
printf("enter element to be searched:");
scanf("%d",&b);
for(i=0;i<10;i++)
{
if(a[i]==b)
printf("element found at location %d\n",i);
}
}

Answer / anshul

#include<stdio.h>
void main()
{
int a[10],i,b;
for(i=0;i<10;i++)
{scanf("%d",&a[i]);
}
printf("enter element to be searched:");
scanf("%d",&b);
for(i=0;i<10;i++)
{
if(a[i]==b)
printf("element found at location %d\n",i);
}
}
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