Практикум по программированию. Основы. Ветвление. Корни уравнения
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Revision as of 08:26, 15 March 2016
Contents[hide] |
Java
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 | 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++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 | // 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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