PAT B1034 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

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 = 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("
");

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

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

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

注意点：这道题的坑有两个，一个是int不够大，两个大int乘起来就爆了，要用long long，long long 的输入输出要用lld。第二个是约分，求最大公约数，遍历就超时了，要用辗转相除法，一定要记住！！！