C Write a program that replies either Leap Year or Not a Lea
Solution
\"1. Leap year or not\"
#include <iostream>
using namespace std;
int main()
{
int year;
cout << \"Enter a year: \";
cin >> year;
if (year % 4 == 0)
{
if (year % 100 == 0)
{
if (year % 400 == 0)
cout << year << \" is a leap year.\";
else
cout << year << \" is not a leap year.\";
}
else
cout << year << \" is a leap year.\";
}
else
cout << year << \" is not a leap year.\";
return 0;
}
\"2. Prime number or not\"
#include <iostream>
#include <cctype>
using namespace std;
int main(){
int n,i=1,sum=0;
cout << \"Enter a number: \";
cin >> n;
cout<<\" Divisors of number are : \";
while(i<n){
if(n%i==0)
{
cout<<i<< \" \";
sum=sum+i;
}
i++;
}
if(sum==n)
cout << endl<<n<<\" is a perfect number\ \";
else
cout << i << \" is not a perfect number\ \";
system(\"pause\");
return 0;
}
\"3. \"
#include <iostream>
#include <cctype>
using namespace std;
int main(){
int low,high,num,sum=0, check=0,cube=0;
cout << \"Enter a lower limit: \";
cin >> low;
cout << \"Enter a higher limit : \";
cin >> high;
cout<<\" calculating.... \"<<endl;
while(low<=high){
num=low;
while ( num > 0 ) {
cube=0;
//cout<<num<< \" \";
sum = num % 10;
cube += (sum * (sum*sum)) ; //calculating cube of each digit
num /= 10;
check+=cube; //adding sum of cubes of digits
}
if (low==check) //checkng sum of cube of digits to number
cout << low <<\" ==> sum is \"<< check<< \"==>\"<< low<<\" is equal to sum of cube of its digits\"<<endl;
check=0;
low++;
}
return 0;
}

