题目描述
- 本题要求编写程序,计算2个有理数的和、差、积、商。
输入描述:
- 输入在一行中按照“a1/b1 a2/b2”的格式给出两个分数形式的有理数,
- 其中分子和分母全是整型范围内的整数,负号只可能出现在分子前,分母不为0。
输出描述:
- 分别在4行中按照“有理数1 运算符 有理数2 = 结果”的格式顺序输出2个有理数的和、差、积、商。
- 注意输出的每个有理数必须是该有理数的最简形式“k a/b”,其中k是整数部分,a/b是最简分数部分;
- 若为负数,则须加括号;
- 若除法分母为0,则输出“Inf”。
- 题目保证正确的输出中没有超过整型范围的整数。
输入例子:
- 5/3 0/6
- 2/3 -4/2
输出例子:
- 1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf
- 2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
思路
- 分解分数
- 将分数形式的字符串,分解成整数+最简真分数形式的字符串
- 整数部分为分子除以分母的值
- 真分数部分的分子需要求余运算,若整体不是最简,需要同时除以最大公约数化为最简
- 输出注意是否有0
- 输出注意负数加括号
- 四则运算
- 四则运算的参数使用最初的分数形式
- 左操作数的分子分母和右操作数的分子分母
- 按分数的四则规则运算,并返回整数+最简真分数形式的字符串
- 除法若除数的分子为0,返回Inf
代码实现
package com.liuyong666.pat;import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);String left = sc.next();String right = sc.next();int zi1 = Integer.parseInt(left.split("/")[0]);int mu1 = Integer.parseInt(left.split("/")[1]);int zi2 = Integer.parseInt(right.split("/")[0]);int mu2 = Integer.parseInt(right.split("/")[1]);System.out.println(fenjieNum(left) + " + " + fenjieNum(right) + " = " +add(zi1,mu1,zi2,mu2));System.out.println(fenjieNum(left) + " - " + fenjieNum(right) + " = " +sub(zi1,mu1,zi2,mu2));System.out.println(fenjieNum(left) + " * " + fenjieNum(right) + " = " +mul(zi1,mu1,zi2,mu2));System.out.println(fenjieNum(left) + " / " + fenjieNum(right) + " = " +div(zi1,mu1,zi2,mu2));}//分解一个分数形式的有理数,返回表示有理数的字符串public static String fenjieNum(String left){StringBuilder newLeft = new StringBuilder();int left_zi = Integer.parseInt(left.split("/")[0]);int left_mu = Integer.parseInt(left.split("/")[1]);int left_zhengshu = left_zi / left_mu;int left_fenshu_zi = left_zi % left_mu;int left_fenshu_gcd = gcd(left_fenshu_zi,left_mu);int left_new_zi = left_fenshu_zi / left_fenshu_gcd;int left_new_mu = left_mu / left_fenshu_gcd;if(left_zhengshu != 0 && left_fenshu_zi != 0){if(left_zi * left_mu < 0){newLeft.append("(-");}newLeft.append(Math.abs(left_zhengshu));newLeft.append(" ");newLeft.append(Math.abs(left_new_zi) + "/" + Math.abs(left_new_mu));if(left_zi * left_mu < 0){newLeft.append(")");}}else if(left_zhengshu != 0){if(left_zi * left_mu < 0){newLeft.append("(");}newLeft.append(left_zhengshu);if(left_zi * left_mu < 0){newLeft.append(")");}}else if(left_fenshu_zi != 0){if(left_zi * left_mu < 0){newLeft.append("(-");}newLeft.append(Math.abs(left_new_zi) + "/" + Math.abs(left_new_mu));if(left_zi * left_mu < 0){newLeft.append(")");}}else{newLeft.append("0");}return newLeft.toString();}//加法 5/3 0/6这种形式的俩字符串为参数public static String add(int zi1, int mu1, int zi2, int mu2){int new_zi = zi1 * mu2 + zi2 * mu1;int new_mu = mu1 * mu2;return fenjieNum(new_zi + "/" + new_mu);}//减法public static String sub(int zi1, int mu1, int zi2, int mu2){int new_zi = zi1 * mu2 - zi2 * mu1;int new_mu = mu1 * mu2;return fenjieNum(new_zi + "/" + new_mu);}//乘法public static String mul(int zi1, int mu1, int zi2, int mu2){return fenjieNum(zi1 * zi2 + "/" + mu1 * mu2);}//除法public static String div(int zi1, int mu1, int zi2, int mu2){if(zi2 == 0){return "Inf";}return mul(zi1, mu1, mu2, zi2);}//最大公约数public static int gcd(int x, int y){return y == 0 ? x : gcd(y, x % y);}}
