蓝桥网试题 java 基础练习矩形面积交

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概述

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思路见锦囊2

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算法

 1 import java.util.Scanner;
 2 public class Main {
 3
public static void main(String[] args) {
 4
Scanner sc = new Scanner(System.in);
 5
double[] x = new double[4];
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double[] y = new double[4];
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double[] z = new double[4];
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for(int i=0;i<4;i++){
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x[i] = sc.nextDouble();
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y[i] = sc.nextDouble();
11 
}
12
z[0] = Math.max(Math.min(x[0], x[1]),Math.min(x[2], x[3])); //x1
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z[1] = Math.min(Math.max(x[0], x[1]),Math.max(x[2], x[3])); //x2
14
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z[2] = Math.max(Math.min(y[0], y[1]),Math.min(y[2], y[3])); //y1
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z[3] = Math.min(Math.max(y[0], y[1]),Math.max(y[2], y[3])); //y2
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if(z[0]<z[1]&&z[2]<z[3])
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System.out.printf("%.2fn",(z[1]-z[0])*(z[3]-z[2]));
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else
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System.out.println("0.00");
21 
}
22 }