标签:lse string 教程 else alt bad 简单 存在 不能
统计题目完成数并计算正确率
设计测试类,利用JUnit测试整数类与分数类的四则运算
public Fraction (int numer, int denom)
方法,以及Math.abs(numerator)
来保证算法不出现漏洞。除此之外还要注意分子分母计算的一些限定特性。我们同样加进代码里。
源代码及分析注释:
import java.util.Random;
public class Fraction {
private int numerator, denominator; //定义分母、分子
public Fraction (int numer, int denom)
{
if(denom == 0 ) //分子为0
denom = 1;
if (denom < 0) //若分母小于0,则取分母为正值,分子为负值
{
numer = numer * -1;
denom = denom * -1;
}
numerator = numer;
denominator = denom;
reduce();
}
public int getNumerator()
{
return numerator;
}
public int getDenominator()
{
return denominator;
}
public Fraction add(Fraction op2) //实现真分数加法运算
{
int commonDenominator = denominator * op2.getDenominator(); //两随机数a1、a2的分母相乘,进行通分
int numerator1 = numerator * op2.getDenominator(); //a1的分子=a1的分子与a2的分母相乘
int numerator2 = op2.getNumerator() * denominator; //a2的分子=a2的分子与a1的分母相乘
int sum = numerator1 + numerator2; //将通分过后的两个随机数相加
System.out.print("("+this.toString()+")" + " + " + "("+op2.toString()+")" + "=");
return new Fraction (sum, commonDenominator);
}
public Fraction subtract(Fraction op2) //实现真分数减法运算
{
int commonDenominator = denominator * op2.getDenominator();
int numerator1 = numerator * op2.getDenominator();
int numerator2 = op2.getNumerator() * denominator;
int difference = numerator1 - numerator2;
System.out.print("("+this.toString()+")" + " - " + "("+op2.toString()+")" + "=");
return new Fraction(difference,commonDenominator);
}
public Fraction multiply (Fraction op2) //实现真分数乘法运算
{
int numer = numerator * op2.getNumerator();
int denom = denominator * op2.getDenominator();
System.out.print("("+this.toString()+")" + " * " + "("+op2.toString()+")" + "=");
return new Fraction (numer, denom);
}
public Fraction divide (Fraction op2) //实现真分数除法运算
{
int numer = numerator * op2.getDenominator();
int denom = denominator * op2.getNumerator();
System.out.print("("+this.toString()+")" + " / " + "("+op2.toString()+")" + "=");
return new Fraction (numer, denom);
}
public String toString() //输出格式及限定
{
String result;
if (numerator == 0) //分子为0,结果为0
result = "0";
else
if(denominator == 0) //分母不能为0
return "错误!分母不能为0";
else
if (denominator == 1) //分母为1,结果取分子值
result = numerator + "";
else
result = numerator + "/" + denominator; //按分数形式输出结果
return result;
}
private void reduce()
{
if (numerator != 0)
{
int common = gcd (Math.abs(numerator), denominator); //取分子分母最大公因子
numerator = numerator / common; //约分
denominator = denominator / common;
}
}
private int gcd (int num1, int num2) //计算最大公因子
{
if(num2==0)
return num1;
else
return gcd(num2,num1%num2);
}
public static Fraction obj(){ //生成随机数
Random ran = new Random();
return new Fraction(ran.nextInt(100),ran.nextInt(100));
}
}
步骤 | 耗时 | 百分比 |
---|---|---|
需求分析 | 20min | 16% |
设计 | 30min | 24% |
代码实现 | 40min | 32% |
测试 | 20min | 16% |
分析总结 | 15min | 12% |
标签:lse string 教程 else alt bad 简单 存在 不能
原文地址:https://www.cnblogs.com/besty-zyx/p/8910233.html