Display powers in C++ loop [closed] - c++

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I have to solve following exercise
Write a program to display 1,5,25,125 upto n terms.
I am in 11th grade and I have tried of many ways of writing this program.
Value of control variable is one, and it's less than n.
But by how much should it differ so that it obeys the above question?
Please answer if you could in simple language.
Also should I use a special variable for power?
Thanks in advance, Abhijith

Answer 1

Print out the old value times five, starting with 1
Basic mockup:
auto PrintExercise(std::size_t terms) -> void {
std::size_t lastResult = 1;
for (std::size_t i = 0; i < terms; ++i) {
std::cout << std::to_string(lastResult) << std::endl;
lastResult *= 5;
}
}
Edit: Turns out I overthought this. It would be easier to just print the power of the control variable.
auto PrintExercise(std::size_t terms) -> void {
for (std::size_t i = 0; i < terms; ++i) {
std::cout << std::to_string(pow(5,n)) << std::endl;
}
}

Answer 2

Since the correct answers have already been provided, here's the same approach using recursion instead of iteration (loops) with (hopefully) enough comments to explain the process. Just for completeness. Give it a try, it's fun!
#include <iostream>
//value = the value that will be printed
//end = after how many iterations you want to stop
void PowerOfFive( const int value, const int end )
{
//Print the current value to the console. This is more or
//less everything the function does...
std::cout << value << ", ";
//... but a function can also call itself, with slightly different
//values in this case. We decrement "end" by 1 and let the whole
//process stop after "end" reaches 0. As long as we're doing that,
//we're multiplying "value" by five each time.
if ( end != 0 )
{
PowerOfFive( value * 5, end - 1 );
}
}
int main()
{
//Example for the above
//Start:
// 1st PowerOfFive(1, 3)
// --> prints 1
// --> calls 2nd PowerOfFive(1 * 5, 3 - 1)
// --> prints 5
// --> calls 3rd PowerOfFive(5 * 5, 2 - 1)
// --> prints 25
// --> calls 4th PowerOfFive(25 * 5, 1 - 1)
// --> prints 125
// --> function 4 ends because "end" has reached 0
// --> function 3 ends
// --> function 2 ends
// --> function 1 ends
PowerOfFive( 1, 3 );
getchar( );
return 0;
}

Answer 3

It seems you want to print powers of 5 upto n, not sure what you mean by control variable. So this should work
for (int i=0;i<=n;++i) cout << pow(5,i) << ", " ;

Answer 4

Iteration value is by 5, it can be done with pow() function & also by using simple for loop like this.
power=0;
cout<<power;
for(i=0;i<n;i++)
{
power=power*5; // OR power*=5
}
cout<<power;

Answer 5

Im adding code see if it helps
#include<iostream>
#include <cmath>
using namespace std;
int main()
{
int exp;
float base;
cout << "Enter base and exponent respectively: ";
cin >> base >> exp;
for(int i=0;i<exp;i++)
{
cout << "Result = " << pow(base, i);
}
return 0;
}
You have to pass the base and exponent value and for your question it should be base=5 and exp=3 and your output will be till 1 , 5 ,25