Практикум по программированию. Основы. Ветвление. Корни уравнения

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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;
}


Python

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# Python 3
 
import math
 
print('\n Поиск корней уравнения A * X^2 + B * X + C = 0')
print('------------------------------------------------')
 
a = float(input(' Введите A: '))
b = float(input(' Введите B: '))
c = float(input(' Введите C: '))
 
print('------------------------------------------------')
 
d = b * b - 4 * a * c
 
if d > 0:
 
    x1 = (-b + math.sqrt(d)) / (2 * a)
    x2 = (-b - math.sqrt(d)) / (2 * a)
 
    print(' X1 = %.3f' % x1)
    print(' X2 = %.3f' % x2)
 
elif d == 0:
 
    x1 = -b / (2 * a)
 
    print(' X = %.3f' % x1)
 
else:
 
    print(' Уравнение корней не имеет')


Pascal

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JavaScript

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<html lang="ru">
<head>
    <meta charset="UTF-8">
    <script>
        function calc() {
            var a = document.getElementById("aId").value;
            var b = document.getElementById("bId").value;
            var c = document.getElementById("cId").value;
 
            var d = b * b - 4 * a * c;
            var x1, x2;
            var result = "";
 
            if (d > 0) {
 
                x1 = (-b + Math.sqrt(d)) / (2 * a);
                x2 = (-b - Math.sqrt(d)) / (2 * a);
 
                result = "X1 = " + x1.toFixed(3) + "<br>" +
                         "X2 = " + x2.toFixed(3);
 
            } else if (d == 0) {
 
                x1 = -b / (2 * a);
 
                result = "X = " + x1.toFixed(3);
 
            } else {
 
                result = "Уравнение корней не имеет";
            }
 
            document.getElementById("resultId").innerHTML = result;
        }
    </script>
</head>
<body>
 
<p>Поиск корней уравнения A * X^2 + B * X + C = 0</p>
<hr>
<p>Введите A: <input id="aId" size="5"></p>
<p>Введите B: <input id="bId" size="5"></p>
<p>Введите C: <input id="cId" size="5"></p>
<hr>
<p id="resultId"></p>
 
<button onclick="calc()">Рассчитать</button>
 
</body>
</html>



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