- What is the time complexity of the following codes? А. void fun(int n, int x) { for (int i = 1; i < n; i = i * x) //or for(int i = n; i >=1; i = i / x) System.out.println("Hello Lab2"); } %3D В. void fun(int n) if (n < 5) System.out.println("Hello Lab2"); else { for (int i = 0; i

- What is the time complexity of the following codes? А. void fun(int n, int x) { for (int i = 1; i < n; i = i * x) //or for(int i = n; i >=1; i = i / x) System.out.println("Hello Lab2"); } %3D В. void fun(int n) if (n < 5) System.out.println("Hello Lab2"); else { for (int i = 0; i

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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int const MULTIPLIER = 5; is a valid way to declare a constant integer variable. Group of answer choices True False ------- Given the following code segment, what is output to the screen? int num1 = 6; int num2 = 4 * num1++; cout << "num1=" << num1 << " num2=" << num2; Group of answer choices num1=7 num2=24 num1=6 num2=24 num1=6 num2=28 Nothing, because the code does not compile.
Create the generated machine code from snippet code as follow: int i=10, j, x;for (a=10; a>0; a--) {if (a % 2 == 1) i += a;else i /= 2;j = i*2;}x = i+j; please solve max in 30 minutes thank u
Java code Ever wondered how change is disbursed to customers in certain stores? In country X the following notes exists 200,100,50,20,10 and following coins: 5, 1, 50c, 10c, 5c. For the sake of this exercises let us only consider change below 50 see below example: Sample run 1: Enter item name: Apples Enter QTY of Apples: 2 Enter price per item N$: 3.25 Amount tendered N$: 50 Your change is: N$ 43.50 Disbursed as follows: 2 x N$20; 0 x N$10; 0 x N$5; 3 x N$1; 1 x 50c; 0 x 10c; 0 x 5c [Hint: make use of % and / operators to determine the values needed]
Compare the complexities of the following two pieces of code using the cyclomatic complexity metrics. What conclusions can you draw about relative complexity of the code? [6 marks] Code A Code B int i = 1; int j = 1; while (i <= 5) { int i = 2; Playacard(i); if (j =i); if (playhaswon(i)) j++; else i++; i++; else System.out.println(j); i--; i++; }
Im facing problem with this code. #include<stdio.h> #include<math.h> void putnqueen(int); int okqueen(int, int); int setqueen(int); void queen(); char p[10][10]; int n; int main() { int a, b; clrscr(); printf(" Enter the no. of queens : "); scanf("%d", &n); for (a = 0; a < n; ++a) for (b = 0; b < n; ++b) p[a][b] = '.'; putnqueen(0); getch(); return (0); } void putnqueen(int x) { int a; if (x < n) { for (a = 0; a < n; ++a) { if (okqueen(x, a)) { p[x][a] = 'Q'; putnqueen(x + 1); p[x][a] = '.'; } } } else { printf("\n The solution is "); queen();…
C++ Given an integer num, repeatedly add all its digits until the result has only one digit, and return it. Example 1: Input: num = 38 Output: 2 Explanation: The process is 38 --> 3 + 8 --> 11 11 --> 1 + 1 --> 2 Since 2 has only one digit, return it. Example 2: Input: num = 0 Output: 0 Constraints: 0 <= num <= 231 - 1 Follow up: Could you do it without any loop/recursion in O(1) runtime? //code here class Solution {public: int addDigits(int num) { }}; //
[Very Easy] Given base and n that are both 1 or more, compute recursively (no loops) the value of base to the n power, so powerN(3, 2) is 9 (3 squared). powerN(3, 1) → 3 powerN(3, 2) → 9 powerN(3, 3) → 27 [Easy] class Trace: def hMB(self,h): if (h==0): print("Stop: ",h) return 0 elif(h==1): print("Stop: ",h) return h else: print("Continue: ",h) return h + self.hMB(h-1) #Tester t = Trace() print("Finally ",t.hMB(5)) [Medium] Suppose, you have been given a non-negative integer which is the height of a ‘house of cards’. To build such a house you at-least require 8 cards. To increase the level (or height) of that house, you would require four sides and a base for each level. Therefore, for the top level, you would require 8 cards and for each of the rest of the levels below you would require 5 extra cards. If you were asked to build level one only, you would require just 8 cards. Of course, the input can be zero; in that case, you do not build a house at all. Complete…
ISP-Java Write a program whose input is two integers, and whose output is the first integer and subsequent increments of 10 as long as the value is less than or equal to the second integer. For coding simplicity, output a space after every integer, including the last. Ex: If the input is: -15 30 the output is: -15 -5 5 15 25 Ex: If the second integer is less than the first as in: 20 5 the output is: Second integer can't be less than the first.
JAVA Question 2: For two integers m and n, their GCD (Greatest Common Divisor) can be computed by a recursive method. Write a recursive method gcd(m,n) to find their Greatest Common Divisor. Method body: If m is 0, the method returns n. If n is 0, the method returns m. If neither is 0, the method can recursively calculate the Greatest Common Divisor with two smaller parameters: One is n, the second one is m mod n (or m % n). The recursive method cannot have loops. Note: although there are other approaches to calculate Greatest Common Divisor, please follow the instructions in this question, otherwise you will not get the credit. main method: Prompt and read in two numbers to find the greatest common divisor. Call the gcd method with the two numbers as its argument. Print the result to the monitor. Example program run: Enter m: 12 Enter n: 28 GCD(12,28) = 4 And here is what I have so far, package CSCI1302;import java.util.*;public class RecursionDemo { public static void…
Identify all private, shared variables in the following code snippet. What is the output of the program (e.g. the value of IS)? int A[100]; int count; void work(int index[]) { float B[10]; …… } void main(){ int IS = 10; int i=0; #pragma omp parallel private(i) { int index[10]; work(index); #pragma omp for reduction(+:IS) for(i=0;i<100;i++) { IS = IS + i; } } printf(“%i\n”, IS); }
Problem 4: Magic Square Test A magic square of order n is an arrangement of n × n numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. See all rows’, columns’ and diagonal’s sum is same. Write a main method where first you need to input n. Then input n*n integers and form a 2D matrix. Print “YES” if the matrix is magic square and print “NO” Otherwise. Sample input: 3 2 7 6 9 5 1 4 3 8 Output: YES
1. Analyze the time complexity of the code segment and find their Big-O.void myfunction1(int n){for(int i=0; i < n; i++) {for(int j=0; j < n; j++) {for(int k=0; k < n; k++) {for(int m=0; m < n; m++) {printf("Hello!");} } } }}