Answer:
while True:
number = int(input("Enter a number: "))
product = number * 10
if product > 100:
break
print(str(product))
Explanation:
Create a while loop that iterates until a specific condition is created inside
Ask the user for the input
Multiply the input and put the result in product
Check if the product is greater than 100. If it is, stop the loop using break keyword
When the loop is done, print the product
You should follow A. Alternate between single and horizontal line breaks, and B. update your resume on job sites once a year, so that you can tell the people who look at the resume what skills you have updated on.
Answer:
import java.util.Arrays;
import java.util.Scanner;
public class num4 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("How many numbers? ");
int n = in.nextInt();
int []intArray = new int[n];
//Entering the values
for(int i=0; i<intArray.length;i++){
System.out.println("Enter the numbers");
intArray[i]=in.nextInt();
}
System.out.println(Arrays.toString(intArray));
int min =intArray[0];
for(int i =0; i<intArray.length; i++){
if(min>intArray[i]){
min = intArray[i];
}
}
System.out.println("The Minimum of the numbers is "+min);
}
}
Explanation:
- Using Java programming language
- Prompt the user for the number of values
- Using Scanner class receive and store in a variable
- Create an array of size n
- Using an for loop continuously ask the user to enter the integers
- Print the array of integers
- Using another for loop with an if statement, find the smallest element in the array of numbers
- Output the the smallest number
Answer:
Explanation:
Algorithm:
a. In each day, you will have to loop through the hotels that come to the hotel after you stayed last night.
b. If a hotel 'h' is found at more than 'd' distance away from last stayed hotel, then the hotel previous of 'h' is chosen to wait for that night. This is the greedy step, and you stay in this hotel.
c. The process for steps a and b is then repeated until we've reached the last hotel xn.
Running time:
Notice that the worst case occurs if each hotel is at a distance of successive multiples of 'd'. The best move is to estimate the distance to each hotel twice the whole computation in the scenario.
Thus, the total running time that could occur in the worst case is O(2n) = O(n). This is said to be linear time.
