蓝桥杯:P1103

解析： 题目不难，主要是考察复数的四则运算公式，这玩意高中学的早就忘记了，有点坑。另外题目要求自己封装四则运算的方法。

package _3_5_test;

import java.util.Scanner;

/*P1103
 * */
public class SeventyOne {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);

		String sign = scanner.next();
		double a = scanner.nextDouble();
		double b = scanner.nextDouble();
		double c = scanner.nextDouble();
		double d = scanner.nextDouble();

		double re[] = null;

		if (sign.equals("+")) {
			re = add(a, b, c, d);
		} else if (sign.equals("-")) {
			re = sub(a, b, c, d);
		} else if (sign.equals("*")) {
			re = mul(a, b, c, d);
		} else if (sign.equals("/")) {
			re = div(a, b, c, d);
		}

		System.out.printf("%.2f+%.2fi", re[0], re[1]);

	}

	public static double[] add(Double a, Double b, Double c, Double d) {
		double result[] = new double[2];
		result[0] = a + c;
		result[1] = b + d;
		return result;
	}

	public static double[] sub(Double a, Double b, Double c, Double d) {
		double result[] = new double[2];
		result[0] = a - c;
		result[1] = b - d;
		return result;
	}

	// 复数的乘法公式：(a+bi)(c+di)=ac+adi+bci-bd
	public static double[] mul(Double a, Double b, Double c, Double d) {
		double result[] = new double[2];
		result[0] = (a * c - b * d);
		result[1] = (a * d + b * c);
		return result;
	}

	// 复数的加法公式：(a+bi)/(c+di)=((ac+bd)+(bc-ad)i)/(c*c+d*d)
	public static double[] div(Double a, Double b, Double c, Double d) {
		double result[] = new double[2];
		result[0] = (a * c + b * d) / (c * c + d * d);
		result[1] = (b * c - a * d) / (c * c + d * d);
		return result;
	}

}