### 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#includeusing namespace std;//Function powerfloat 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");}