Question

Question 4: Set P to the largest digit in your 5 digit number. For example, if your 5 digit number is 88412, then P should be set to 8. Calculate and print the value of (P+1) multiplied by itself (P+3) times. As an example, if your 5 digit number is 88412, you should print the result of multiplying 9 with itself 12 times -- this is also known as 9 to the power of 12. HINT: you are free to re-arrange the code any way you see fit HINT: you are free to add more variables if needed. HINT: you are free to use known math functions HINT: alternatively, you are free to use for loops write a c++ code

Answer 1

#include <iostream>
using namespace std;

// Maximum number of digits in
// output
#define MAX 100000

// This function multiplies x
// with the number represented by res[].
// res_size is size of res[] or
// number of digits in the number
// represented by res[]. This function
// uses simple school mathematics
// for multiplication.
// This function may value of res_size
// and returns the new value of res_size
int multiply(int x, int res[], int res_size) {

   // Initialize carry
   int carry = 0;

   // One by one multiply n with
   // individual digits of res[]
   for (int i = 0; i < res_size; i++) {
       int prod = res[i] * x + carry;

       // Store last digit of
       // 'prod' in res[]
       res[i] = prod % 10;

       // Put rest in carry
       carry = prod / 10;
   }

   // Put carry in res and
   // increase result size
   while (carry) {
       res[res_size] = carry % 10;
       carry = carry / 10;
       res_size++;
   }
   return res_size;
}

// This function finds
// power of a number x
void power(int x, int n)
{

   //printing value "1" for power = 0
   if (n == 0) {
       cout << "1";
       return;
   }

   int res[MAX];
   int res_size = 0;
   int temp = x;

   // Initialize result
   while (temp != 0) {
       res[res_size++] = temp % 10;
       temp = temp / 10;
   }

   // Multiply x n times
   // (x^n = x*x*x....n times)
   for (int i = 2; i <= n; i++)
       res_size = multiply(x, res, res_size);

   cout << x << "^" << n << " = ";
   for (int i = res_size - 1; i >= 0; i--)
       cout << res[i];
}

int findP(int num) {
   int max = -1;
   int mod;
   while (num != 0) {
       mod = num % 10;
       num = num / 10;
       if (max < mod) max = mod;
   }
   return max;
}

// Driver program
int main() {
   int exponent = 100;
   int base = 20;
   cout << "Enter 5 digit number: "<<endl;
   int num;
   cin >> num;
   int p = findP(num);
//cout << "Maximum :" << p << endl;
   power(p+1, p+3);
   return 0;
}