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; }