1088 Rational Arithmetic(20 分)
For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient.
Input Specification:
Each input file contains one test case, which gives in one line the two rational numbers in the format a1/b1 a2/b2. The numerators and the denominators are all in the range of long int. If there is a negative sign, it must appear only in front of the numerator. The denominators are guaranteed to be non-zero numbers.
Output Specification:
For each test case, print in 4 lines the sum, difference, product and quotient of the two rational numbers, respectively. The format of each line is number1 operator number2 = result. Notice that all the rational numbers must be in their simplest form k a/b, where k is the integer part, and a/b is the simplest fraction part. If the number is negative, it must be included in a pair of parentheses. If the denominator in the division is zero, output Inf as the result. It is guaranteed that all the output integers are in the range of long int.
Sample Input 1:
2/3 -4/2
Sample Output 1:
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
Sample Input 2:
5/3 0/6
Sample Output 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
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
struct Fraction{
ll up,down;
}a,b; ll gcd(ll a,ll b){
return b == ? a : gcd(b,a%b);
} Fraction reduction(Fraction result){
if(result.down < ){
result.down = -result.down;
result.up = -result.up;
}
if(result.up == ){
result.down = ;
}else{
int d = gcd(abs(result.up),abs(result.down));
result.down /= d;
result.up /= d;
}
return result;
} Fraction add(Fraction f1,Fraction f2){
Fraction result;
result.up = f1.up*f2.down + f1.down*f2.up;
result.down = f1.down*f2.down;
return reduction(result);
} Fraction muli(Fraction f1,Fraction f2){
Fraction result;
result.up = f1.up*f2.down - f1.down*f2.up;
result.down = f1.down*f2.down;
return reduction(result);
} Fraction multi(Fraction f1,Fraction f2){
Fraction result;
result.up = f1.up*f2.up;
result.down = f1.down*f2.down;
return reduction(result);
} Fraction divide(Fraction f1,Fraction f2){
Fraction result;
result.up = f1.up * f2.down;
result.down = f1.down * f2.up;
return reduction(result);
} void showResult(Fraction r){
r = reduction(r);
if(r.up < ) printf("(");
if(r.down == ) printf("%lld",r.up);
else if(abs(r.up) > r.down){
printf("%lld %lld/%lld",r.up/r.down,abs(r.up)%r.down,r.down);
}else{
printf("%lld/%lld",r.up,r.down);
}
if(r.up < ) printf(")");
} int main(){
scanf("%lld/%lld %lld/%lld",&a.up,&a.down,&b.up,&b.down); showResult(a);
printf(" + ");
showResult(b);
printf(" = ");
showResult(add(a,b));
printf("\n"); showResult(a);
printf(" - ");
showResult(b);
printf(" = ");
showResult(muli(a,b));
printf("\n"); showResult(a);
printf(" * ");
showResult(b);
printf(" = ");
showResult(multi(a,b));
printf("\n"); showResult(a);
printf(" / ");
showResult(b);
printf(" = ");
if(b.up == ) printf("Inf");
else showResult(divide(a,b)); return ;
}
1088 Rational Arithmetic(20 分)的更多相关文章
- 【PAT甲级】1088 Rational Arithmetic (20 分)
题意: 输入两个分数(分子分母各为一个整数中间用'/'分隔),输出它们的四则运算表达式.小数需要用"("和")"括起来,分母为0的话输出"Inf&qu ...
- PAT Advanced 1088 Rational Arithmetic (20) [数学问题-分数的四则运算]
题目 For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate ...
- PAT甲题题解-1088. Rational Arithmetic (20)-模拟分数计算
输入为两个分数,让你计算+,-,*,\四种结果,并且输出对应的式子,分数要按带分数的格式k a/b输出如果为负数,则带分数两边要有括号如果除数为0,则式子中的结果输出Inf模拟题最好自己动手实现,考验 ...
- PAT (Advanced Level) 1088. Rational Arithmetic (20)
简单题. 注意:读入的分数可能不是最简的.输出时也需要转换成最简. #include<cstdio> #include<cstring> #include<cmath&g ...
- 1088. Rational Arithmetic (20)
1.注意在数字和string转化过程中,需要考虑数字不是只有一位的,如300转为"300",一开始卡在里这里, 测试用例: 24/8 100/10 24/11 300/11 2.该 ...
- pat1088. Rational Arithmetic (20)
1088. Rational Arithmetic (20) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue F ...
- PAT 1088 Rational Arithmetic[模拟分数的加减乘除][难]
1088 Rational Arithmetic(20 分) For two rational numbers, your task is to implement the basic arithme ...
- PAT 1088 三人行(20 分)(暴力破解+流程分析)
1088 三人行(20 分) 子曰:"三人行,必有我师焉.择其善者而从之,其不善者而改之." 本题给定甲.乙.丙三个人的能力值关系为:甲的能力值确定是 2 位正整数:把甲的能力值的 ...
- PAT Rational Arithmetic (20)
题目描写叙述 For two rational numbers, your task is to implement the basic arithmetics, that is, to calcul ...
随机推荐
- 在Debug中使用断点调试程序
我最近在学习汇编的程序,所以很多都需要动手写点代码去测试,如果是测试三五行代码的还比较简单,可以在debug中直接按T进行单步调试,但是到后来调试的代码越来越复杂,越来越长,如果再使用单步调试不知道要 ...
- socket发送结构体
struct send_info {char info_from[20]; //发送者IDchar info_to[20]; //接收者IDint info_length; //发送的消息主体的长度c ...
- mysqllog
-- mysql delete log online 1 mysql命令purge mysql> purge master logs to "mysql-bin.000410&quo ...
- Hdu 4762 网络赛 高精度大数模板+概率
注意题目中的这句话he put the strawberries on the cake randomly one by one,第一次选择草莓其实有N个可能,以某一个草莓为开头,然后顺序的随机摆放, ...
- AI:AI
ylbtech-AI:AI 人工智能(Artificial Intelligence),英文缩写为AI.它是研究.开发用于模拟.延伸和扩展人的智能的理论.方法.技术及应用系统的一门新的技术科学. 人工 ...
- JSP标签和EL表达式
1.jsp标签: sun原生的,直接jsp使用 <jsp:include> -- 实现页面包含,动态包含 <jsp:include page="/index.jsp&quo ...
- 问题:HttpWebRequest request post 传参; 结果:好用的C# HttpWebRequest用Post同时提交参数和文件的封装类
在项目中,本来都是在站内进行数据交互的,后来又加进来一个买的php网站(艹).需要进行数据交互,在没有考虑使用web服务的情况下,只有通过Post提交到页面进行数据交互是最好的方式了. 我这边使用的是 ...
- Jmeter测试接口简单使用教程
1. 打开 解决 apache-jmeter-2.13 然后进解压后的然后点击bin 文件里面的jmeter.bat 打开jmeter 2. 添加测试组件 1:添 ...
- Servlet编程实例 续3
----------------siwuxie095 Servlet 跳转之请求的转发 修改 LoginServlet.java: package com.siwuxie095.servlet; im ...
- 关于android上dpi/screen-size的厘清解释
android定义了四种screen-size: small normal large xlarge 同时定义了六种dpi级别: ldpi (low) ~120dpimdpi (medium) ~16 ...