Практикум по программированию. Основы. Ветвление. Корни уравнения

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Revision as of 08:26, 15 March 2016

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Contents

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Java

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import java.util.Scanner;
import java.lang.Math;
 
public class Task01 {
 
    public static void main(String[] args) {
 
        Scanner in = new Scanner(System.in);
 
        System.out.println();
        System.out.println(" Поиск корней уравнения A * X^2 + B * X + C = 0");
        System.out.println("------------------------------------------------");
 
        System.out.print(" Введите A: ");
        double a = in.nextDouble();
 
        System.out.print(" Введите B: ");
        double b = in.nextDouble();
 
        System.out.print(" Введите C: ");
        double c = in.nextDouble();
 
        System.out.println("------------------------------------------------");
 
        double d = b * b - 4 * a * c;
        double x1, x2;
 
        if (d > 0) {
 
            x1 = (-b + Math.sqrt(d)) / (2 * a);
            x2 = (-b - Math.sqrt(d)) / (2 * a);
 
            System.out.printf(" X1 = %.3f\n", x1);
            System.out.printf(" X2 = %.3f\n", x2);
 
        } else if (d == 0) {
 
            x1 = -b / (2 * a);
 
            System.out.printf(" X = %.3f\n", x1);
 
        } else {
 
            System.out.println(" Уравнение корней не имеет");
 
        }
    }
}


C++

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// g++ 4.2
 
#include <iostream>
#include <math.h>
 
using namespace std;
 
int main() {
     
    double a, b, c;
     
    cout << "\n Поиск корней уравнения A * X^2 + B * X + C = 0\n";
    cout << "------------------------------------------------\n";
     
    cout << " Введите A: ";
    cin >> a;
     
    cout << " Введите B: ";
    cin >> b;
     
    cout << " Введите C: ";
    cin >> c;
     
    cout << "------------------------------------------------\n";
     
    double d = b * b - 4 * a * c;
    double x1, x2;
     
    if (d > 0) {
 
        x1 = (-b + sqrt(d)) / (2 * a);
        x2 = (-b - sqrt(d)) / (2 * a);
         
        printf(" X1 = %.3f\n", x1);
        printf(" X2 = %.3f\n", x2);
         
    } else if (d == 0) {
             
        x1 = -b / (2 * a);
             
        printf(" X = %.3f\n", x1);
             
    } else {
             
        cout << " Уравнение корней не имеет\n";
             
    }
     
    cout << "\n";
    return 0;
}


Pascal

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JavaScript

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