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Commit 2c449e0

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Merge pull request fnplus#655 from annu12340/master
Created new files (factorial, armstrong-number, recursion/palindrome)
2 parents 4e580b2 + e7d71c6 commit 2c449e0

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7 files changed

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7 files changed

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#include <iostream>
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using namespace std;
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int main()
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{
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int i,fact=1,number;
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cout<<"Enter any Number: ";
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cin>>number;
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for(i=1;i<=number;i++){
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fact=fact*i;
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}
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cout<<"Factorial of " <<number<<" is: "<<fact<<endl;
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return 0;
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}
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num = int(input("Enter a number: "))
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factorial = 1
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# check if the number is negative, positive or zero
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if num < 0:
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print("Sorry, factorial does not exist for negative numbers")
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elif num == 0:
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print("The factorial of 0 is 1")
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else:
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for i in range(1,num + 1):
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factorial = factorial*i
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print("The factorial of",num,"is",factorial)
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"""
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Or you can directly use the math.factorial() in python
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import math
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num = int(input("Enter a number: "))
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print ("The factorial is ",math.factorial(num), end="")
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"""
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## Armstrong number
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- Armstrong number is a number that is equal to the sum of cubes of its digits.
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Example: 0, 1, 153, 370, 371 , 407
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#include <iostream>
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using namespace std;
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int main()
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{
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int n,r,sum=0,temp;
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cout<<"Enter the Number= ";
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cin>>n;
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temp=n;
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while(n>0)
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{
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r=n%10;
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sum=sum+(r*r*r);
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n=n/10;
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}
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if(temp==sum)
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cout<<"Armstrong Number."<<endl;
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else
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cout<<"Not Armstrong Number."<<endl;
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return 0;
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}
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#include <bits/stdc++.h>
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using namespace std;
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int rev(int n, int temp)
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{
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if (n == 0)
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return temp;
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// stores the reverse of a number
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temp = (temp * 10) + (n % 10);
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return rev(n / 10, temp);
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}
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int main()
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{
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cout<<"Enter the Number= ";
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cin>>n;
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int temp = rev(n, 0);
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if (temp == n)
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cout << "yes" << endl;
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else
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cout << "no" << endl;
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return 0;
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}
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import java.io.*;
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class Palindrome
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{
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static int rev(int n, int temp)
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{
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if (n == 0)
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return temp;
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// stores the reverse of a number
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temp = (temp * 10) + (n % 10);
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return rev(n / 10, temp);
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}
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public static void main (String[] args)
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{
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int n = 101;
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int temp = rev(n, 0);
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if (temp == n)
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System.out.println("yes");
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else
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System.out.println("no" );
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}
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}
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def rev(n, temp):
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if (n == 0):
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return temp;
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# stores the reverse of a number
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temp = (temp * 10) + (n % 10);
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return rev(n / 10, temp);
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n = int(input("enter a number : "));
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temp = rev(n, 0);
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if (temp != n):
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print("yes");
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else:
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print("no");
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