Power to be able
Write a program that receives two integers m and n from the user andOnly with the operator of the pluralThe number m is to the power of n
Write a program that receives two integers m and n from the user andOnly with the operator of the pluralThe number m is to the power of n
let msg = 'error' alert(msg)This answer is only visible to premium members
This answer is only visible to premium members
let msg = 'error' alert(msg)This answer is only visible to premium members
This answer is only visible to premium members
let msg = 'error' alert(msg)This answer is only visible to premium members
This answer is only visible to premium members
m = int(input("please enter a number: ")) n = int(input("please enter a number: ")) a = 1 for i in range(0, n): a *= m print(a)
Where is the collector ????
def power_using_addition(m, n): # If n is negative, the result is 0 if n < 0: return 0 # If n is equal to 0, the result is 1 elif n == 0: return 1 result = 0 for _ in range(n): result += m # The sum of m to the number of n return result # Get input from the user try: m = int(input("عدد m را وارد کنید: ")) n = int(input("عدد n را وارد کنید: ")) # The calculation and display of the result result = power_using_addition(m, n) print(f"{m} به توان {n} برابر است با: {result}") except ValueError: print("لطفاً یک عدد صحیح وارد کنید.")
def click(m,n):
return f"natije = {m ** n}"
print(click(m=int(input("enter:")),n=int(input("enter:"))))
/*
این بخش رو فقط برای پیدا کردن یک فرمول برای حل این سوال نوشتم
در ضمن زبان برنامه سی پلاس پلاس هست
x = شماره الگو = n - 1
y = تعداد ضرب در خودش = n * m
n = عدد اول = input
m = عدد دوم = input
x n ** m ==> y = n * m ==>
---------------------------------------------------------------------
1 : 3 ** 2 ==> 3 * 3 ==> 3 + 3 + 3
2 : 3 ** 3 ==> 3 * 3 * 3 ==> 3 + 3 + 3 + 3 + 3 + 3
3 : 3 ** 4 ==> 3 * 3 * 3 * 3 ==> 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3
0 = 1 = 2 = 3 = 4 = 5 = 6
1 = 3 = 9 = 27 = 81 = 243 = 729
*/
#include <conio.h>
#include <iostream>
using namespace std;
int main() {
int x, y;
while (1 == 1) {
cout<< " Please enter x and y or enter 0 and 0 for exit : " ;
cin >> x >> y;
int l = 1;
int i = 1;
if (x != 0 && y != 0) {
while (i <= y) {
int temp = 0;
for (int j = 0; j < x; ++j) {
temp += l;
}
l = temp;
i++;
}
}
else {
cout << '\n' << '\n' << " OK. Thanks for use my app, Bye!!" ;
break;
}
cout<< " It is Result : " ;
cout << " " << l << '\n' << '\n' ;
}
getch();
}
def tavan(m,n): a=0 if n==0: result=1 print(result) elif n>m: print('n bayad az m kamtar basheh!') # I can't just solve this part with the collector for now else: for i in range(1,m+1): a+=m print(a) m=int(input('m: ')) n=int(input('n: ')) tavan(m,n)
n = int(input("n value: ")) m = int(input("m value: ")) def add(a, b): num = a for i in range(b): num += 1 return num def multiply(a, b): num = 0 for i in range(b): num = add(num, a) return num def exponent(a, b): num = 1 for i in range(b): num = multiply(num, a) return num print(exponent(n, m))
def taha(n , m):
return pow(n , m)
x = taha( int(input ("enter number1: ")) , int(input ("enter number2 : ")) )
print(x)
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