1034 有理数四则运算 （20 分）

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<algorithm>
using namespace std;

typedef long long ll;

struct Fraction{
    ll up,down;
}a,b; 

ll gcd(ll a, ll b){
    return b == 0?a:gcd(b,a%b);
}

Fraction reduction(Fraction result){
    if(result.down < 0){
        result.down = -result.down;
        result.up = -result.up;
    }
    if(result.up == 0) result.down = 1;
    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 + f1.down * f2.up;
    result.down = f1.down*f2.down;
    return result;
}

Fraction minu(Fraction f1,Fraction f2){
    Fraction result;
    result.up = f1.up*f2.down - f1.down * f2.up;
    result.down = f1.down*f2.down;
    return result;
}

Fraction multi(Fraction f1,Fraction f2){
    Fraction result;
    result.up = f1.up*f2.up;
    result.down = f1.down*f2.down;
    return result;
}

Fraction divide(Fraction f1,Fraction f2){
    Fraction result;
    result.up = f1.up*f2.down;
    result.down = f1.down*f2.up;
    return 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,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",&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(minu(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 == 0) printf("Inf");
    else showResult(divide(a,b));
    printf("\n");
    
    return 0;
}