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
就一句话,细节很重要!
 #include <iostream>
using namespace std;
long long dv, res1, res2;
long long DIV(long long a, long long b)
{
if (b == )
return abs(a);
return DIV(b, a%b);
}
void print(long long a1, long long b1, long long a2, long long b2, char c)
{
if (a1 == )
printf("%d %c ", , c);
else
{
printf("%s", a1 > ? "" : "(");
dv = DIV(a1, b1);
a1 /= dv;
b1 /= dv; if (a1 / b1 != )
printf("%d", a1 / b1);
if (a1 - b1 * (a1 / b1) != )
printf("%s%d/%d", a1 / b1 != ? " " : "", a1 / b1 != ? abs(a1 - b1 * (a1 / b1)) : a1, b1);
printf("%s %c ", a1 > ? "" : ")", c);
} if(a2 == )
printf("%d %s ", , "=");
else
{
printf("%s", a2 > ? "" : "(");
dv = DIV(a2, b2);
a2 /= dv;
b2 /= dv;
if (a2 / b2 != )
printf("%d", a2 / b2);
if (a2 - b2 * (a2 / b2) != )
printf("%s%d/%d", a2 / b2 != ? " " : "", a2 / b2 != ? abs(a2 - b2 * (a2 / b2)) : a2, b2);
printf("%s %s ", a2 > ? "" : ")", "=");
} if (res1 == )
{
printf("%d\n",);
return;
}
else if (res2 == )
{
printf("Inf\n");
return;
}
printf("%s", res1 > ? "" : "(");
dv = DIV(res1, res2);
res1 /= dv;
res2 /= dv;
if (res1 / res2 != )
printf("%d", res1 / res2);
if (res1 - res2 * (res1 / res2) != )
printf("%s%d/%d", res1 / res2 != ? " " : "", res1 / res2 != ? abs(res1 - res2 * (res1 / res2)) : res1, res2);
printf("%s\n", res1 > ? "" : ")");
}
int main()
{
char c;
long long a1, b1, a2, b2;
cin >> a1 >> c >> b1 >> a2 >> c >> b2;
// +
res1 = a1 * b2 + a2 * b1;
res2 = b1 * b2;
print(a1, b1, a2, b2, '+');
// -
res1 = a1 * b2 - a2 * b1;
res2 = b1 * b2;
print(a1, b1, a2, b2, '-');
// *
res1 = a1 * a2;
res2 = b1 * b2;
print(a1, b1, a2, b2, '*');
// /
res1 = a2 > ? a1 * b2 : a1 * b2*-;
res2 = b1 * abs(a2);
print(a1, b1, a2, b2, '/');
return ;
}

PAT甲级——A1088 Rational Arithmetic的更多相关文章

  1. A1088. Rational Arithmetic

    For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate the ...

  2. PAT 甲级 1081 Rational Sum (数据不严谨 点名批评)

    https://pintia.cn/problem-sets/994805342720868352/problems/994805386161274880 Given N rational numbe ...

  3. PAT甲级——A1081 Rational Sum

    Given N rational numbers in the form numerator/denominator, you are supposed to calculate their sum. ...

  4. 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 ...

  5. PAT_A1088#Rational Arithmetic

    Source: PAT A1088 Rational Arithmetic (20 分) Description: For two rational numbers, your task is to ...

  6. 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 ...

  7. PAT甲级题解(慢慢刷中)

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给 ...

  8. PAT甲级题分类汇编——计算

    本文为PAT甲级分类汇编系列文章. 计算类,指以数学运算为主或为背景的题. 题号 标题 分数 大意 1058 A+B in Hogwarts 20 特殊进制加法 1059 Prime Factors ...

  9. PAT甲级题分类汇编——序言

    今天开个坑,分类整理PAT甲级题目(https://pintia.cn/problem-sets/994805342720868352/problems/type/7)中1051~1100部分.语言是 ...

随机推荐

  1. hashmap1.7的死锁模拟

    package com.cxy.springdataredis.hashmap; import javax.lang.model.element.VariableElement; import jav ...

  2. Redis问题整理

    Redis问题总结 1.单点登录的两个项目cookie不一致 由于在配置自定义Cookie的时候 @Bean("shiroCookie") public SimpleCookie ...

  3. php多维数组排序方案。按照姓名 首字符 等排序

    //定义一个学生数组   $students = array(     256=>array('name'=>'jon','grade'=>98.5),     2=>arra ...

  4. hdu6089 Rikka with Terrorist

    题意:n*m的平面内有K个不安全点,Q个询问位置在(x,y)的人能走到多少个点?走到:(x,y)和(x',y')之间的矩形中不包含不安全点. 标程: #include<bits/stdc++.h ...

  5. Liunx下安装Oracle11g时Oracle Grid安装包下载向导

    下载Oracel 11g  Grid的安装包 Oracle官网 https://www.oracle.com 快捷访问路径:https://www.oracle.com/technetwork/dat ...

  6. WPF命令好状态刷新机制

    https://blog.csdn.net/WPwalter/article/details/90344470 this.DispatcherInvoke(() => { System.Wind ...

  7. ios 中倒计时计算,时间戳为NaN

    // 倒计时 daojishi(params) { let _this = this; let datetemp = this.servertimes; let lasttime = Date.par ...

  8. DELPHI实现类似仿360桌面的程序界面

    1.窗体半透明: Alphablend属性为true;Alphablendvalue的值为100 2.窗体透明: formCreate: Self.TransparentColor := True;S ...

  9. Activity详解一 配置、启动和关闭activity转载 https://www.cnblogs.com/androidWuYou/p/5887726.html

    先看效果图: Android为我们提供了四种应组件,分别为Activity.Service.Broadcast receivers和Content providers,这些组建也就是我们开发一个And ...

  10. HDU-3068-最长回文-马拉车算法模板题

    给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度. 回文就是正反读都是一样的字符串,如aba, abba等 Input输入有多组case,不超过120组,每组输入为 ...