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How to find the power of a number with only addition and loops?

There is a question in my assessment to write a program that takes 2 numbers from the user n1 and n2 and finds n1 to the power n2 (n1^n2). I am not allowed to use the math library or simple multiplication. All I can use is simple addition and loops. I tried a lot but can't get the program right. Please help.

2nd Nov 2018, 11:36 AM
Razin Bhatti
7 odpowiedzi
+ 9
You can also use the Web version of SoloLearn.
2nd Nov 2018, 4:35 PM
blACk sh4d0w
blACk sh4d0w - avatar
+ 8
Welcome 😊
2nd Nov 2018, 5:10 PM
blACk sh4d0w
blACk sh4d0w - avatar
+ 7
Like others said, if you are having trouble with your code, post a link to your code so that you can get helpful answers and others can identify your mistakes which will help you more in learning rather than asking for a code. :) #include <iostream> using namespace std; int Multiplication(int num, int sum); int main() { int num = 0, pow = 0, sum = 0; cin >> num; cin >> pow; cout << "You entered " << num << "^" << pow << endl; for(int i = 0; i < pow - 1; i++) { sum = Multiplication(num, sum); } cout << num << "^" << pow << " = " << sum << endl; return 0; } int Multiplication(int num, int sum) { int temp = (sum == 0)? num : sum; sum = 0; for(int i = 0; i < num; i++) sum += temp; return sum; }
2nd Nov 2018, 2:34 PM
blACk sh4d0w
blACk sh4d0w - avatar
+ 2
Show your attempt to receive some helpful answers.
2nd Nov 2018, 12:15 PM
MsJ
MsJ - avatar
+ 1
You could begin by making your own multiplication function. Multiplying a number x by y means nothing more than adding x to itself y - 1 times. This is something you can achieve using addition and a for-loop. Using this function and the same pattern you can afterwards design your power function. Again, raising a number x to y means multiplying x by itself y - 1 times. Here you can use a for-loop again. For more detailed help, you should provide us with your own attempt just as Mohit suggested so we can see what you are trying to do.
2nd Nov 2018, 12:36 PM
Shadow
Shadow - avatar
+ 1
Thanks alot nAutAxh AhmAd. I code on my laptop so that's why I couldn't post my code.
2nd Nov 2018, 4:30 PM
Razin Bhatti
0
Okay I'll give it a try. Thanks for your help☺
2nd Nov 2018, 4:40 PM
Razin Bhatti