本题要求编写程序,计算 2 个有理数的和、差、积、商。
输入格式:
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。
输出格式:
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。
输入样例 1:
2/3 -4/2
输出样例 1:
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
输入样例 2:
5/3 0/6
输出样例 2:
1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf
Think
做除法运算时先要判断分母。
import java.io.*; public class Main { //辗转相除法 private static long GCD(long a,long b) { return b == 0 ? a : GCD(b , a % b); } //相加 private static String calculate(long a,long b) { if(b == 0) { return "Inf"; } long gcd,t,x; gcd = GCD(Math.abs(a), b); //最大公约数 //简化 a = a / gcd; b = b / gcd; t = Math.abs(a) / b; //整数 x = Math.abs(a) - t * b;//分子 if(t == 0 && x == 0) { return "0"; } if(a < 0) { if(t != 0 && x != 0) return "(-"+t+" "+x+"/"+b+")"; if(t != 0 && x == 0) return "(-"+t+")"; if(t == 0 && x != 0) return "(-"+x+"/"+b+")"; } else { if (t != 0 && x != 0) return t+" "+x+"/"+b; if(t != 0 && x == 0) return String.valueOf(t); if(t == 0 && x != 0) return x+"/"+b; } return null; } public static void main(String[] args) throws IOException { BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); String[] istr = in.readLine().split(" "); String[] a = istr[0].split("/"), b = istr[1].split("/"); long a1 = Long.parseLong(a[0]), a2 = Long.parseLong(b[0]); long b1 = Long.parseLong(a[1]), b2 = Long.parseLong(b[1]); String A,B; A = calculate(a1,b1); B = calculate(a2,b2); out.println(A + " + " + B + " = " + calculate(a1*b2+a2*b1,b1*b2)); out.flush(); out.println(A + " - " + B + " = " + calculate(a1*b2-a2*b1,b1*b2)); out.flush(); out.println(A + " * " + B + " = " + calculate(a1*a2,b1*b2)); out.flush(); out.print(A + " / " + B + " = "); out.flush(); if(a2 < 0) { out.print(calculate(a1 * b2 * a2 / Math.abs(a2) , b1 * Math.abs(a2))); } else { out.print(calculate(a1 * b2 , b1 * a2)); } out.flush(); } }