PAT乙级1034

Posted 2022-08-30 wx630c98f24f6b8

1034. 有理数四则运算(20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct fraction

  long long  a, b;
  long long k;
  fraction():k(0)
;
int gcd(long long a, long long b)

  return !b ? a : gcd(b, a%b);

fraction reduction(fraction f)

  fraction ft;
  if (f.b < 0)
  
    ft.a = -f.a;
    ft.b = -f.b;
    f.a = -f.a;
    f.b = -f.b;
  
   if (f.a == 0)
  
    ft.a = 0;
    ft.b = 1;
    f.a = 0;
    f.b = 1;
  
  
  ft.a= f.a/gcd(abs(f.a), f.b);
  ft.b= f.b/gcd(abs(f.a), f.b);
  if (abs(ft.a) >= ft.b)
  
    if (f.a < 0)
    
      ft.k = -abs(ft.a) / ft.b;
      ft.a = abs(ft.a) % ft.b;
    
    else
    
      ft.k = abs(ft.a) / ft.b;
      ft.a = abs(ft.a) % ft.b;
    
  
  return ft;

fraction add(fraction f1, fraction f2)

  fraction f;
  f.a = f1.a*f2.b + f2.a*f1.b;
  f.b = f1.b*f2.b;
  return reduction(f);

fraction substract(fraction f1, fraction f2)

  fraction f;
  f.a = f1.a*f2.b - f2.a*f1.b;
  f.b = f1.b*f2.b;
  return reduction(f);

fraction multiply(fraction f1, fraction f2)

  fraction f;
  f.a = f1.a*f2.a;
  f.b = f1.b*f2.b;
  return reduction(f);

fraction divide(fraction f1, fraction f2)

  fraction f;
  f.a = f1.a*f2.b;
  f.b = f1.b*f2.a;
  return reduction(f);

void printFraction(fraction f)

  if (f.k > 0)
  
    if (f.a != 0)
    
      cout << f.k << " " << f.a << "/" << f.b;
    
    else
      cout << f.k;
  
  else if (f.k == 0)
  
    if (f.a > 0)
    
      cout << f.a << "/" << f.b;
    
    else if (f.a == 0)
      cout << 0;
    else
      cout << "(" << f.a << "/" << f.b << ")";
  
  else
  
    if (f.a != 0)
      cout << "(" << f.k << " " << f.a << "/" << f.b<<")";
    else
      cout << "(" << f.k << ")";
  

int main()

  fraction f1, f2;
  scanf("%lld/%lld %lld/%lld", &f1.a, &f1.b, &f2.a, &f2.b);
  printFraction(reduction(f1));
  cout << " + ";
  printFraction(reduction(f2));
  cout << " = ";
  printFraction(add(f1, f2));
  cout << endl;
  printFraction(reduction(f1));
  cout << " - ";
  printFraction(reduction(f2));
  cout << " = ";
  printFraction(substract(f1, f2));
  cout << endl;
  printFraction(reduction(f1));
  cout << " * ";
  printFraction(reduction(f2));
  cout << " = ";
  printFraction(multiply(f1, f2));  
  cout << endl;
  printFraction(reduction(f1));
  cout << " / ";
  printFraction(reduction(f2));
  cout << " = ";
  if (f2.a != 0)
    printFraction(divide(f1, f2));
  else
    cout << "Inf";
  cout << endl;
  return 0;