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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

#include <iostream>

#include <stdio.h>

#include <algorithm>

#include <string>

#include <map>

using namespace std;

const int maxn = ;

struct num{

    long long k = ;

    long long up;

    long long down;

};

int gcd(long long a, long long b){

    return b ==  ? 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 < ){

        res.down = - * res.down;

        res.up = - * res.up;

    }

    return res;

}

void clean(num n){

    int flag = ;

    n.k = n.up / n.down;

    if (n.up < ){

        n.up = -n.up;

        flag = ;

    }

    n.up = n.up%n.down;

    int g = abs(gcd(n.up, n.down));

    n.up /= g;

    n.down /= g;

    if (flag == ){

        printf("(");

    }

    if (n.k != ){

        printf("%lld", n.k);

        if (n.up != ){

            printf(" %lld/%lld", n.up, n.down);

        }

    }

    else{

        if (n.up != ){

            if (flag == )printf("-");

            printf("%lld/%lld", n.up, n.down);

        }

        else{

            printf("");

        }

    }

    if (flag == ){

        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 != ){

        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");

}