Answered: #include int k=0; int f(int t) { k +=…

#include int k=0; int f(int t) { k += t; heturn k; } void main() int i, result=1; { for(i = 1;i <= 3;i++) result = f(i); printf("%d\n", result); } A 3 B C 1 D 123

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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PROGRAMMING LANGUAGE: C++ SUBJECT: OOP Task: Find errors in the following code and rewrite a correct version of code with outputs. #include <iostream> using namespace std; class CRectangle { int width. height; public: void set_values (int, int); int area () {return (width * height);} }; void CRectangle::set_values (int a, int b) { width = a; height = b; } int main () { CRectangle a,*b,*c; c= a; a.set_values (1,2); b= new CRectangle; c=&b; c->set_values (3,4); cout << "a area: " << c->area() << endl;// cout << "*b area: " << a.area() << endl;// c->set_values (5,6); cout << "*b area: " << b.area() << endl;// delete b; return 0; }
Python question Analysis: Computational Complexity (Q14-15) In this section, you are asked to analyse the worst-case computational time complexity of Python functions. Always identify the order of growth in the tightest and simplest possible form using big-O notation, e.g., write O(n) instead of O(2n+n/2). Question 14 (Complexity 1) State the overall worst-case time complexity of the below function and provide a short explanation of your answer. The function accepts an integer n as input. def mystery(n): num = 1 count = 0 while num < n: num = num*2 count += 1 return count Question 15 (Complexity 2) State the overall worst-case time complexity of the below function and provide a short explanation of your answer. The function is supposed to return a list, res, containing the duplicated elements in the given list , lst, of size n. Analyse the function in terms of the length of the list, n. def duplicate_values(lst): res = [] for i in range(len(lst)):…
int pertambahan(int pilihan, int now){ int result; result =pilihan + now; return result;} int perkalian(int pilihan, int now){ int result; result = pilihan*now; return result;} int main(){ int NoR,initial; int now; int temp1, temp2; scanf("%d %d", &NoR, &initial); getchar(); now=initial; for(int o=0;o<NoR;o++){ char a,b; int pilihan1,pilihan2; scanf("%c %d %c %d",&a, &pilihan1,&b, &pilihan2); getchar(); if(a=='+') { if(pilihan1>=-1000 && pilihan1<=1000) { temp1 = pertambahan(pilihan1,now); } else{ temp1=-10000; } } else if(a=='x') { if(pilihan1>=-2 && pilihan1<=2){ temp1 = perkalian(pilihan1,now); } else { temp1=-10000; } } if(b=='+') { if(pilihan2>=-1000 && pilihan2<=1000) { temp2 = pertambahan(pilihan2,now); } else {…
For the following code #include <stdio.h> #include <conio.h> int powerof(int base, int pow); int _tmain(int argc, _TCHAR* argv[]) { int x,y,z; line 3 called
For the following main function int _tmain(int argc, _TCHAR* argv[]) { int years[10]={2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016}; cout<<"Years before updating: "<
#include #include #include struct Point2d { double x; double y; void print() { // example: (2.5,3.64) std::cout << "(" << x << "," << y << ")" << std::endl; } double length() const { return std::sqrt(x * x + y * y); } // const Point2D & ==> function can't modify 'other' // the second const ==> function can't modify x, y Point2d add(const Point2d &other) const { Point2d result; result.x = x + other.x; result.y = y + other.y; return result; } }; // // do NOT modify above this line // Point2d discplacement(std::string commands) { // TODO: re-write this Point2d p = {0, 0}; return p; } // // do NOT modify below this line // void check(Point2d result, double x , double y) { if (result.x != x || result.y != y) { std::cout << "fail" << std::endl; } else { std::cout << "pass" << std::endl; } } int main() { Point2d result =…
Please explain the code below #include <cstdio> #include <math.h> using namespace std; int main ( ) { // radius const int RAD = 10; // tolerance const int TOL = 5; for(int x = -RAD; x <= RAD; x++){ for(int y = -RAD; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " "); printf(abs(eq) < TOL ? "*" : " "); } for(int y = -RAD+4; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " "); printf(abs(eq) < TOL ? "*" : " "); } for(int y = -RAD+4; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " "); printf(abs(eq) < TOL ? "*" : " "); } for(int y = -RAD+4; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " ");…
Programming language: Java
C programming #include <stdio.h>int main() {int i, j, n ;printf("height? ") ;scanf("%2d", &n) ;for (i = 1 ; i <= n ; i++) {// printf("%d: ", i) ;for (j = 1 ; j <= i ; j++) {// Pick *one* of the following// printf("%d", j % 10) ;printf("*") ;}printf("\n") ;}return 0 ;} How can I get output like this by editing the given programming -- height? 5************************* and then how can I get the following?- height? 7 1 123 12345 1234567 123456789 12345678901 1234567890123
#include using namespace std; void some_action_1(int); int main() { cout<
#include <stdio.h>#include <stdlib.h> typedef struct Number_struct { int num;} Number; void Swap(Number* numPtr1, Number* numPtr2) { /* Your code goes here */} int main(void) { Number* num1 = NULL; Number* num2 = NULL; num1 = (Number*)malloc(sizeof(Number)); num2 = (Number*)malloc(sizeof(Number)); int int1; int int2; scanf("%d", &int1); scanf("%d", &int2); Thank you so much but i forgot to put this how i would fit in into this. num1->num = int1; num2->num = int2; Swap(num1, num2); printf("num1 = %d, num2 = %d\n", num1->num, num2->num); return 0;}
Int cube ( int x){ // returns cube of x : return x.x.x. :)* O False True