本題要求編寫程序，計算 2 個有理數的和、差、積、商。

輸入格式：

輸入在一行中按照 a1/b1 a2/b2 的格式給出兩個分數形式的有理數，其中分子和分母全是整型範圍內的整數，負號只可能出現在分子前，分母不為 0。

輸出格式：

分別在 4 行中按照 有理數1 運算符 有理數2 = 結果 的格式順序輸出 2 個有理數的和、差、積、商。註意輸出的每個有理數必須是該有理數的最簡形式 k a/b，其中 k 是整數部分，a/b 是最簡分數部分；若為負數，則須加括號；若除法分母為 0，則輸出 Inf。題目保證正確的輸出中沒有超過整型範圍的整數。

輸入樣例 1：

2/3 -4/2

輸出樣例 1：

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
 

輸入樣例 2：

5/3 0/6

輸出樣例 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 <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <map>
using namespace std;
const int maxn = 100010
 
;
struct num{
    long long k = 0;
    long long up;
    long long down;
};
int gcd(long long a, long long b){
    return b == 0 ? a : gcd(b, a % b);
}
num add(num n1, num n2){
    num res;
    res.down = n1.down*n2.down;
    res.up = n1.down*n2.up + n1.up*n2.down;
    return res;
}
num sub(num n1, num n2){
    num res;
    res.down 
 
= n1.down*n2.down;
    res.up = n2.down*n1.up - n2.up*n1.down;
    return res;
}
num mul(num n1, num n2){
    num res;
    res.down = n1.down*n2.down;
    res.up = n1.up*n2.up;
    return res;
}
num dive(num n1, num n2){
    num res;
    res.down = n1.down*n2.up;
    res.up = n1.up*n2.down;
    if (res.down < 0){
        res.down = -1 * res.down;
        res.up = -1 * res.up;
    }
    return res;
}
void clean(num n){
    int flag = 0;
    n.k = n.up / n.down;
    if (n.up < 0){
        n.up = -n.up; 
        flag = 1;
    }
    n.up = n.up%n.down;
    int g = abs(gcd(n.up, n.down));
    n.up /= g;
    n.down /= g;
    if (flag == 1){
        printf("(");
    }
    if (n.k != 0){
        printf("%lld", n.k);
        if (n.up != 0){
            printf(" %lld/%lld", n.up, n.down);
        }
    }
    else{
        if (n.up != 0){
            if (flag == 1)printf("-");
            printf("%lld/%lld", n.up, n.down);
        }
        else{
            printf("0");
        }
    }
    if (flag == 1){
        printf(")");
    }

}
int main(){
    num n1, n2;
    scanf("%lld/%lld %lld/%lld", &n1.up, &n1.down, &n2.up, &n2.down);
    /*int g = abs(gcd(n1.up, n1.down));
    n1.up /= g;
    n1.down /= g;
    g = abs(gcd(n2.up, n2.down));
    n2.up /= g;
    n2.down /= g;*/
    num q, w, e, r;
    q = add(n1, n2);
    clean(n1);
    printf(" + ");
    clean(n2);
    printf(" = ");
    clean(q);
    printf("\n");

    w = sub(n1, n2);
    clean(n1);
    printf(" - ");
    clean(n2);
    printf(" = ");
    clean(w);
    printf("\n");

    r = mul(n1, n2);
    clean(n1);
    printf(" * ");
    clean(n2);
    printf(" = ");
    clean(r);
    printf("\n");

    if (n2.up != 0){
        e = dive(n1, n2);
        clean(n1);
        printf(" / ");
        clean(n2);
        printf(" = ");
        clean(e);
        printf("\n");
    }
    else{
        clean(n1);
        printf(" / ");
        clean(n2);
        printf(" = Inf\n");
        
    }
    
    system("pause");
}

註意點：這道題的坑有兩個，一個是int不夠大，兩個大int乘起來就爆了，要用long long，long long 的輸入輸出要用lld。第二個是約分，求最大公約數，遍歷就超時了，要用輾轉相除法，一定要記住！！！

PAT B1034 有理數四則運算 （20 分）