PAT B1034

1034. 有理数四则运算(20)

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

代码如下：

#include <cstdio>
#include <cmath>
typedef long long ll;
using namespace std;
struct Fraction{
  ll up, down;
}f1, f2;
ll gcd(ll a, ll b){
  return !b ? a : gcd(b, a%b);
}
Fraction reduction(Fraction result){
  if(result.down < 0){
    result.up = -result.up;
    result.down = -result.down;
  }
  if(result.up == 0){
    result.down = 1;
  } else {
    int d = gcd(abs(result.up), abs(result.down));
    result.up /= d;
    result.down /= d;
  }
  return result;
}
Fraction add(Fraction f1, Fraction f2){
  Fraction result;
  result.up = f1.up * f2.down + f2.up * f1.down;
  result.down = f1.down * f2.down;
  return reduction(result);
}
Fraction minu(Fraction f1, Fraction f2){
  Fraction result;
  result.up = f1.up * f2.down - f2.up * f1.down;
  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 < 0) printf("(");
    if(r.down == 1) {
        printf("%lld", r.up);
    } else if(abs(r.up) > r.down){
        printf("%lld %lld/%lld", r.up / r.down, (ll)abs(r.up) % r.down, r.down);
    } else {
        printf("%lld/%lld", r.up, r.down);
    }
    if(r.up < 0) printf(")");
}
int main(){
  scanf("%lld/%lld %lld/%lld", &f1.up, &f1.down, &f2.up, &f2.down);
  showResult(f1);printf(" + ");showResult(f2);printf(" = ");showResult(add(f1, f2));printf("\n");
  showResult(f1);printf(" - ");showResult(f2);printf(" = ");showResult(minu(f1, f2));printf("\n");
  showResult(f1);printf(" * ");showResult(f2);printf(" = ");showResult(multi(f1, f2));printf("\n");
  showResult(f1);printf(" / ");showResult(f2);printf(" = ");
  if(f2.up == 0) printf("Inf");
  else showResult(divide(f1, f2));
  return 0;
}