Wednesday, 25 May 2016
Write a program that inputs a number form the user and display all perfect number upto the number entered.
#include<iostream.h>
#include <conio.h>
void main()
{
int num , mid, sum, limit;
clrscr();
cout<<"Enter the number upto which perfect numbers are required. ";
cin>>limit;
if(limit < 6)
cout<<"No perfect number upto " <<limit;
else
{
cout<<"Perfect numbers upto "<<limit<< " are "<<endl;
for(num = 6; num <= limit; num++)
{
sum = 0;
mid = num / 2;
for(int i=1; i<= mid; i++)
{
if((num % i)== 0)
sum += i;
}
if(sum == num)
cout<<num << " ";
}
}
getch();
}
Write a program that input a number from the user and display all Armstrong numbers upto the number entered.
#include<iostream.h>
#include<conio.h>
int main()
{
int num, n, sum, r , limit;
clrscr();
cout<<"Enter the number upto which armstrong numbers are required. ";
cin>> limit;
cout<<"Armstrong number upto "<<limit<<" are "<<endl;
for(num = 1; num <= limit; num++)
{
n = num;
sum = 0;
while (n != 0)
{
r = n % 10;
sum += r * r * r;
n /= 10;
}
if(sum == num)
cout<<num<< " ";
}
getch();
}
#include<iostream.h>
#include<conio.h>
int main()
{
int num, n, sum, r , limit;
clrscr();
cout<<"Enter the number upto which armstrong numbers are required. ";
cin>> limit;
cout<<"Armstrong number upto "<<limit<<" are "<<endl;
for(num = 1; num <= limit; num++)
{
n = num;
sum = 0;
while (n != 0)
{
r = n % 10;
sum += r * r * r;
n /= 10;
}
if(sum == num)
cout<<num<< " ";
}
getch();
}
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