乙级 PAT 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

思路：比较繁琐的题目

#include<stdio.h>
#include<math.h>


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

void print(long long a, long long b){
    long long c; 
    if(a > 0){ 
        if(b == 1){ 
            printf("%lld", a);
        }
        else if(a > b){ 
            c = a / b;
            a -= b * c;
            printf("%lld %lld/%lld", c, a, b);
        }
        else{ 
            printf("%lld/%lld", a, b);
        }
    }
    else if(a == 0){ 
        printf("%c", '0');
    }
    else{ 
        if(b == 1){ 
            printf("(%lld)", a);
        }
        else if(-1 * a > b){ 
            c = a / b;
            a = (-1 * a) % b;
            printf("(%lld %lld/%lld)", c, a, b);
        }
        else{ 
            printf("(%lld/%lld)", a, b);
        }
    }
}

void add(long long a, long long b, long long c, long long d){
    print(a, b);
    printf(" + ");
    print(c, d);
    printf(" = ");
    long long m = a * d + c * b;
    long long n = b * d;
    
    long long gcd3 = abs(gcd(m, n));
    m /= gcd3;
    n /= gcd3;
    print(m, n);
    printf("\n");
}

void sub(long long a, long long b, long long c, long long d){
    print(a, b);
    printf(" - ");
    print(c, d);
    printf(" = ");
    long long m = a * d - c * b;
    long long n = b * d;
    
    long long gcd3 = abs(gcd(m, n));
    m /= gcd3;
    n /= gcd3;
    print(m, n);
    printf("\n");
}

void mul(long long a, long long b, long long c, long long d){
    print(a, b);
    printf(" * ");
    print(c, d);
    printf(" = ");
    long long m = a * c;
    long long n = b * d;
    
    long long gcd3 = abs(gcd(m, n));
    m /= gcd3;
    n /= gcd3;
    print(m, n);
    printf("\n");
}

void div(long long a, long long b, long long c, long long d){
    print(a, b);
    printf(" / ");
    print(c, d);
    printf(" = ");
    if(c == 0){
        printf("Inf");
    }
    else if(c < 0){
        long long m = -1 * a * d;
        long long n = -1 * b * c;
        
        long long gcd3 = abs(gcd(m, n));
        m /= gcd3;
        n /= gcd3;
        print(m, n);
    }
    else{
        long long m = a * d;
        long long n = b * c;
        
        long long gcd3 = abs(gcd(m, n));
        m /= gcd3;
        n /= gcd3;
        print(m, n);
    }
    printf("\n");
}

int main(){
    long long a, b, c, d;
    long long c1 = 0, c2 = 0;
    scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
   
    long long gcd1 = abs(gcd(a, b));
    a /= gcd1;
    b /= gcd1;
    long long gcd2 = abs(gcd(c, d));
    c /= gcd2;
    d /= gcd2;
  
    add(a, b, c, d);
    sub(a, b, c, d);
    mul(a, b, c, d);
    div(a, b, c, d);
    return 0;
}