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#include <iostream>using namespace std;main{cout << "No tabbing. That's very sad :(\n";cout << "No in-editor highlighting either :(((\n";cout << "Descriptions might be niice too.";}
#include <iostream>#include <cstring>#include <unistd.h>#include <sys/utsname.h>int main() {char newHostname[] = "newhostname"; // Replace with the desired hostnameif (sethostname(newHostname, strlen(newHostname)) == 0) {std::cout << "Hostname set to: " << newHostname << std::endl;// Optionally, update the /etc/hostname file to make the change permanentFILE *hostnameFile = fopen("/etc/hostname", "w");if (hostnameFile != NULL) {fprintf(hostnameFile, "%s\n", newHostname);fclose(hostnameFile);} else {perror("Failed to update /etc/hostname");}} else {perror("Failed to set hostname");}return 0;}
#include <iostream>#include <fstream>#include <string>#include <cstring>using namespace std;//This program makes a new text file that contains all combinations of two letters.// aa, ab, ..., zy, zzint main(){string filename = "two_letters.txt";ofstream outFile;outFile.open(filename.c_str());if(!outFile.is_open()){cout << "Something's wrong. Closing..." << endl;return 0;}for(char first = 'a'; first <= 'z'; first++){for(char second = 'a'; second <= 'z'; second++){outFile << first << second << " ";}outFile << endl;}return 0;}
#include <iostream>int main(){const char* const hello = "Hello, world!";const char* bruh = hello;char* const yeet = hello;std::cout << bruh << std::endl;std::cout << yeet << std::endl;return 0;}/*Place your bets!Will the program:a.) Print "Hello, world!" twice?b.) Compile error on line 5 (bruh initialize line) because the pointer gets implicit cast to non-const?c.) Compile error on line 7 (yeet initialize line) because the char gets implicit cast to non-const?d.) Both b and c?e.) Compile error line 11 (print yeet) because the pointer is constant and can't be incrementedf.) Print "Hello, world!" then print the pointer address in hexadecimalg.) Both b and e?h.) Both c and e?i.) B, c, and e?*/// The answer is in this base 64 string:// 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
/*Good morning! Here's your coding interview problem for today.This problem was asked by Stripe.Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well.For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.You can modify the input array in-place.*/#include <iostream>using namespace std;int calcMissing(int* input, int size){int sum = 0;int n = 1; //add one to account for missing valuefor(int i = 0; i < size; i++){if(input[i] > 0){sum += input[i];n++;}}//If no numbers higher than 0, answer is 1if(sum == 0)return 1;return (n*(n+1)/2) - sum; //Formula is expectedSum - actualSum/* expectedSum = n*(n+1)/2, the formula for sum(1, n) */}int main(){cout << calcMissing(new int[4]{3, 4, -1, 1}, 4) << endl;cout << calcMissing(new int[3]{1, 2, 0}, 3) << endl;//No positive numberscout << calcMissing(new int[1]{0}, 1) << endl;}
#include <iostream>using namespace std;int main(){cout << "Hello, World!" << endl;return 0;}