Answer:
p(x,n)
1. if(n==0) [if power is 0]
2. then result =1.
3.else
4. { result=1.
5. for i=1 to n.
6. { result = result * x. } [each time we multiply x once]
7. return result.
8. }
Let's count p(3,3)
3
0, so come to else part.
i=1: result = result *3 = 3
i=2: result = result *3 = 9
i=2: result = result *3 = 27
Explanation:
here the for loop at step 4 takes O(n) time and other steps take constant time. So overall time complexity = O(n)
Answer:
Explanation:
The following code is written in Java and loops through 10 times. Each time generating 2 random dice rolls. If the sum is 10 it breaks the loop and outputs a "You Win" statement. Otherwise, it outputs "You Lose"
import java.util.Random;
class Brainly {
public static void main(String[] args) {
UseRandom useRandom = new UseRandom();
boolean youWin = false;
for (int x = 0; x<10; x++) {
int num1 = useRandom.getRandom(6);
int num2 = useRandom.getRandom(6);
if ((num1 + num2) == 10) {
System.out.println("Number 1: " + num1);
System.out.println("Number 2: " + num2);
System.out.println("You Win");
youWin = true;
break;
}
}
if (youWin == false) {
System.out.println("You Lose");
}
}
}
class UseRandom{
public int getRandom(int n)
{
Random r=new Random();
int rand=r.nextInt(n);
return rand;
}}
1. Honeymoon stage
2. Distress and anxiety
3. Adjustment
4. Adaption
5. Re-entry shock
The answer is true because choosing a technology solution is the last step
Answer:
The recursion function is as follows:
def raise_to_power(num, power):
if power == 0:
return 1
elif power == 1:
return num
else:
return (num*raise_to_power(num, power-1))
Explanation:
This defines the function
def raise_to_power(num, power):
If power is 0, this returns 1
if power == 0:
return 1
If power is 1, this returns num
elif power == 1:
return num
If otherwise, it calculates the power recursively
else:
return (num*raise_to_power(num, power-1))