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
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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)
Python
Where is the collector ????
m = int(input("m ra vared konid: "))
n = int(input("n ra vared konid: "))
print(m ** n)
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.")
Python
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();
}
C & C++
from itertools import product
def f01(m , n):
m = m
n = n
m2 = 0
for i in product(range(m),repeat=n-1):
m2 = m2 + m
print(m2)
f01(5,3)
while True:
try:
x = int(input("m: "))
y = input("n: ")
if y == "0":
print("1")
elif x >= 0 and int(y) >= 0:
f01(x,int(y))
else:
print("این برنامه برای محاسبه اعداد منفی نمیباشد")
except ValueError:
print("لطفا عدد وارد کنید")
'''
توضیحات بیشتر:
ماژول پروداکت رو ایمپورت کردم تا به جای نوشتن چند بار حلقه تو در تو از این ماژول استفاده کنم
روش سنتی:
for i in range(m):
for j in range(m):
for h in range(m):
print("hello world")
روش ماژول پروداکت:
for i, j, h in product(range(m),range(m),range(m)):
print("hello world")
برای تعداد تکرار مشخص:
for i in product(range(m),repeat=n):
print("hello world")
'''
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)
Python
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