For instance, if you're doing it in C, it would be something like this:
#include
int main(int argc, char *argv[]){
int answer[5], n;
for(n = 0; n < 5; n++){
printf("Gimme a number: ");
scanf("%i", &answer[n]);
}
for(n = 4; n >= 0; n--){
printf("%i\n", answer[n]);
}
return 0;
}
Add the two numbers and divide that by two.
PLease be more specific.
brobably 4 or 5 numbers mainly they input their dob it means six numbers.
You draw a flowchart to find maximum and minimum of given 3 input numbers by using all three numbers. You take the low, high and input the middle number between them. You can see the rise, or decline of the chart that way.
you people are useless.
To write a C++ program to display the student details using class and array of object.
The time complexity of constructing a segment tree data structure is O(n), where n is the number of elements in the input array. The time complexity of querying a segment tree is O(log n), where n is the number of elements in the input array.
Assuming that the input has already been put into an int[] array, the function to remove duplicates will operate as follows. The function will create an output array that will accept each first unique int of the input array. A comparator will compare each following cell, ignoring duplicates until the end of the array is reached. The output array will then be printed.
Create an array with 50 elements and input the integers one a time, filling the array. Use an insertion sort on the array for each input except the first. Alternatively, input the values first and then use insertion sort.
#include<iostream> #include<sstream> #include<vector> using namespace std; int main() { vector<int> Array; for (size_t index=0; index<9; ++index) { int i; bool valid = false; while (!valid) { cout << "Enter a number: "; string input; cin >> input; stringstream ss; ss << input; valid = (ss >> i); if (!valid) cout << "Invalid input.\n"; } Array.push_back (i); } // print array and average int average=0; for (size_t index=0; index<9; ++index) { cout << Array[index] << ' '; average += Array[index]; } average /= 10; cout << "\nAverage: " << average << endl; }
There are two forms of the NAND gate. Inverted input: Ouput = not Input1 and not Input2 and not Input3 ... Inverted output: Output = not (Input1 and Input2 and Input3 ...)
Logic to search element in array Input size and elements in array from user. ... Input number to search from user in some variable say toSearch . Define a flag variable as found = 0 . ... Run loop from 0 to size . ... Inside loop check if current array element is equal to searched number or not. To learn more about data science please visit- Learnbay.co
An inverting amplifier is one where the output is an inverted function of the input. The Class A transistor amplifier, also known as common-emitter, is inverting. As you increase the voltage on the base, the output voltage on the collector decreases. The operational amplifier has an inverting and a non-inverting input. In typical bridge mode, the output is inverted with respect to the (inverting) input, and the non-inverting input is used to reject common-mode input signals by moving the virtual ground point as needed.
#include "stdio.h" #define ARRAY_SIZE 10 void fill(float* array, int size); void spill(float* array, int size, char* delimiter); void bubble_sort(float* array, int size); void reverse(float* array, int size); void swap(float* a, float* b); int main(int argc, char* argv[]) { float numbers[ARRAY_SIZE]; fill(numbers, ARRAY_SIZE); bubble_sort(numbers, ARRAY_SIZE); spill(numbers, ARRAY_SIZE, " "); reverse(numbers, ARRAY_SIZE); spill(numbers, ARRAY_SIZE, " "); return 0; } void fill(float* array, int size) { int i = 0; while (i < size) fscanf(stdin, "%f", array + (i++)); } void spill(float* array, int size, char* delimiter) { int i = 0; while (i < size) fprintf(stdout, "%f%s", array[i++], delimiter); fputc('\n', stdout); } void bubble_sort(float* array, int size) { int i, j; for (i = 0; i < size; i++) for (j = i; j < size; j++) if (array[i] > array[j]) swap(array + i, array + j); } void reverse(float* array, int size) { int i; for (i = size / 2; i >= 0; i--) swap(array + i, array + (size - (i + 1))); } void swap(float* a, float* b) { float c = *a; *a = *b; *b = c; } fill gets the numbers from input spill sends them to output bubble sort will sort the array in ascending order reverse will reverse the list so that it is in descending order swap is used to swap two floats You can change float to double or int depending on which datatype you want to use.
The time complexity of an algorithm that uses a binary search on a sorted array is O(log n), where n is the size of the input array.
Dendrites.
import java.util.Arrays; import java.util.Scanner; public class Answers { public static void main(String[] args) { //Creates a scanner object named console. Scanner console = new Scanner(System.in); //Variabels int [] numbers = new int [10]; double avg = 0.0; double median = 0.0; int max = numbers[0]; double count = 0.0; //User input. for (int i = 0; i < numbers.length; i++){ System.out.print("Number: "); numbers[i] = console.nextInt(); } //break System.out.println("==============="); //finds the average and max value. for (int i = 0; i < numbers.length; i++){ count += numbers[i]; avg = count / numbers.length; //average if (numbers[i] > max){ //finds the max value. max = numbers[i]; } } median = (numbers[4] + numbers[5])/2; //Median value //Display to user. System.out.println("Highest value found: " + max); //Show maximum value found in array System.out.printf("Median is: %.3f \n",median); //Show median System.out.printf("Average is: %.3f \n",avg); //Show average sortAsc(numbers); //Print out whole array ascending } //Method for sorting an Array ascending. public static void sortAsc(int [] array){ for (int i = 0; i < array.length; i++){ Arrays.sort(array); System.out.println(array[i]); } } } This should do everything you asked for, hope this helps!