• Nov 18, 2022 •AustinLeath
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/* Algorithm: Step 1: Get radius of the cylinder from the user and store in variable r Step 2: Get height of the cylinder from the user and store in variable h Step 3: Multiply radius * radius * height * pi and store in v Step 4: Display the volume */ #include <iostream> using namespace std; int main() { float r; //define variable for radius float h; //define variable for height float v; float pi; pi=3.1416; cout<<"Enter radius:"; cin>>r; cout<<"Enter height:"; cin>>h; v=r*r*h*pi; //compute volume cout<<"Radius:"<<r<<"\tHeight:"<<h<<endl; //display radius and height cout<<"\n************************\n"; cout<<"Volume:"<<v<<endl;//display volume return 0; }
• Aug 5, 2023 •usama
1 like • 6 views
/* 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 value for(int i = 0; i < size; i++) { if(input[i] > 0) { sum += input[i]; n++; } } //If no numbers higher than 0, answer is 1 if(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 numbers cout << calcMissing(new int[1]{0}, 1) << endl; }
• Sep 7, 2022 •LeifMessinger
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#include <iostream> #include <cstring> int main(int argc, char** argv){ //With decimal if(strstr(argv[1], ".") != nullptr){ int i = 0; //Skip i to first non 0 digit while(argv[1][i] < '1' || argv[1][i] > '9') ++i; //If digit comes before decimal if((argv[1] + i) < strstr(argv[1], ".")){ //Good example of pointer arithmetic std::cout << strlen(argv[1] + i) - 1 << std::endl; //Another good example }else{ //If digit is after decimal std::cout << strlen(argv[1] + i) << std::endl; } }else{ //Without decimal int m = 0; int i = 0; while(argv[1][i] < '1' || argv[1][i] > '9') ++i; //In case of some number like 0045 for(; argv[1][i] != '\0'; ++i){ if(argv[1][i] >= '1' && argv[1][i] <= '9') m = i + 1; } std::cout << m << std::endl; } return 0; }
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#include <iostream> using namespace std; /* Description: uses switch case statements to determine whether it is hot or not outside. Also uses toupper() function which forces user input char to be uppercase in order to work for the switch statement */ int main() { char choice; cout << "S = Summer, F = Fall, W = Winter, P = Spring" << endl; cout << "Enter a character to represent a season: ";asdasdasdasd cin >> choice; enum Season {SUMMER='S', FALL='F', WINTER='W', SPRING='P'}; switch(toupper(choice)) // This switch statement compares a character entered with values stored inside of an enum { case SUMMER: cout << "It's very hot outside." << endl; break; case FALL: cout << "It's great weather outside." << endl; break; case WINTER: cout << "It's fairly cold outside." << endl; break; case SPRING: cout << "It's rather warm outside." << endl; break; default: cout << "Wrong choice" << endl; break; } return 0; }
• Apr 15, 2025 •hasnaoui1
0 likes • 4 views
int main()
#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, zz int 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; }