#include using namespace std; void PrintFactorial(int factCounter, int factValue){ int nextCounter; int nextValue; if (factCounter == 0) { // Base case: 0! = 1 cout << "1" << endl; } else if (factCounter == 1) { // Base case: Print 1 and result cout << factCounter << " = " << factValue << endl; } else { // Recursive case cout << factCounter << " * "; nextCounter = factCounter - 1; nextValue = nextCounter * factValue; /* Your solution goes here */ } } int main() { int userVal; cin >> userVal; cout << userVal << "! = "; PrintFactorial(userVal, userVal); return 0; }

Check the explanation

Explanation:

Code in C++::

#include <iostream>

using namespace std;

void PrintFactorial(int factCounter, int factValue){

int nextCounter = 0;

int nextValue = 0;

if (factCounter == 0) { // Base case: 0! = 1

cout << "1" << endl;

}

else if (factCounter == 1) { // Base case: Print 1 and result

cout << factCounter << " = " << factValue << endl;

}

else { // Recursive case

cout << factCounter << " * ";

nextCounter = factCounter - 1;

nextValue = nextCounter * factValue;

/* Your solution goes here */

/**

* We just need to call the function PrintFactorial() recursively

* and pass the two parameters that are just calculated as nextCounter for factCounter

* and nextValue as factValue.

*/

PrintFactorial(nextCounter,nextValue);

}

}

int main() {

int userVal = 0;

userVal = 5;

cout << userVal << "! = ";

PrintFactorial(userVal, userVal);

return 0;

}

Output::

Test Case 1 where userVal=5::

Attached Image 1

Test Case 2 where userVal=6::

Attached Image 1

answer: report them

explanation:

explanation:  

an electrode is a metal whose surface serves as the location where oxidation-reduction equilibrium is established between the metal and what is in the solution.