Program 1

posted Mar 25, 2013, 10:23 PM by Salah Ibrahim   [ updated Mar 26, 2013, 12:12 AM by Karwan Jacksi ]
//Write two c++ functions  one for factorial and another for power,
//then use these two functions to solve this equation : R=1!/n^1 +  3!/n^3 + 5!/n^5 …m!/n^m




#include<iostream>
using namespace std;

//Function power
float pow(int n, int m)            //n,m: are function parameters
{
  float p = 1;                         //p is storage and initialized by 1 because we have multiplication. 
  for(int i = 1; i <= m; i++)   //our loop from 1 -> m. 
     p *= n;                          //multiply n by p and put the result in p.
  return p;                             //return the value of p which is (n to power of m).
}  

//Function factorial               
float fact(int n)                       //n is function parameter
{
  float f = 1;                         //f is storage and initialized by 1 because we have multiplication. 
  for(int i = 1; i <= n; i++)   //our loop from 1 -> n.  
     f *= i;                           //multiply i by f and put the result in f.
  return f;                             //return the value of f which is (factorial of n).
}

void main()        
{
  float sum=0, num1, num2;           //sum is storage and initialized by 0 because we have addition.
  cin >> num1 >> num2;                    //reading num1 then num2 variables.
  for(int i = 1; i <= num1; i+=2)        //our loop from 1 -> num1, its increased by 2 because we only want odd numbers in loop.  
     sum +=  //dividing factorial on power and store result in sum. 
     (fact(i) /                                     //calling function fact(i), i: is function argument. 
     pow(num2, i));                          //calling function pow(num2, i), num2,i: are function arguments.       
  cout << sum << endl;                    //printing sum (the result).
  system("pause");
}
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