PAT甲题题解-1088. Rational Arithmetic (20)-模拟分数计算

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
long long numerator[2];
long long denominator[2];
long long GCD(long long a,long long b){
    a=abs(a);
    b=abs(b);
    if(b==0)
        return a;
    return GCD(b,a%b);

}
struct Fraction{
    long long integer;
    long long numerator;
    long long denominator;
}frac[3];
/**
转化成带分数的形式
*/
void transfer(long long numerator,long long denominator,int i){
    //long long gcd=GCD(numerator,denominator);
    //numerator=numerator/gcd;
    //denominator=denominator/gcd;
    frac[i].integer=numerator/denominator;
    frac[i].numerator=numerator%denominator;
    frac[i].denominator=denominator;
}
/**
输出，如果为负数还要有括号
*/
void print(Fraction f){
    if(f.integer!=0){
        if(f.integer<0)
            printf("(");
        printf("%lld",f.integer);
    }

    if(f.numerator!=0){
        if(f.integer!=0){
            printf(" ");
        }
        //如果前面整数部分为0，则这里需要加(
        else if(f.numerator<0)
            printf("(-");
        //注意，如果是负数，只要整数前面一个‘-‘就可以，所以这里不需要‘-‘
        if(f.numerator>0)
            printf("%lld/%lld",f.numerator,f.denominator);
        else
            printf("%lld/%lld",-f.numerator,f.denominator);
    }
    if(f.integer==0&&f.numerator==0)
        printf("0");
    else if(f.integer<0 || f.numerator<0)
        printf(")"); //末尾补)
}


int main()
{
    long long res;
    scanf("%lld/%lld %lld/%lld",&numerator[0],&denominator[0],&numerator[1],&denominator[1]);
    //先约分
    long long gcd0=GCD(numerator[0],denominator[0]);
    numerator[0]=numerator[0]/gcd0;
    denominator[0]=denominator[0]/gcd0;
    long long gcd1=GCD(numerator[1],denominator[1]);

    numerator[1]=numerator[1]/gcd1;
    denominator[1]=denominator[1]/gcd1;

    long long gcd=GCD(denominator[0],denominator[1]);
    long long lcm=(denominator[0]/gcd)*gcd*(denominator[1]/gcd);
    transfer(numerator[0],denominator[0],0);
    transfer(numerator[1],denominator[1],1);
    //sum
    long long sum=numerator[0]*(lcm/denominator[0])+numerator[1]*(lcm/denominator[1]);
    long long gcd2=abs(GCD(sum,lcm));
    transfer(sum/gcd2,lcm/gcd2,2);
    print(frac[0]);
    printf(" + ");
    print(frac[1]);
    printf(" = ");
    print(frac[2]);
    printf("\n");

    //difference
    sum=numerator[0]*(lcm/denominator[0])-numerator[1]*(lcm/denominator[1]);
    gcd2=abs(GCD(sum,lcm));
    transfer(sum/gcd2,lcm/gcd2,2);
    print(frac[0]);
    printf(" - ");
    print(frac[1]);
    printf(" = ");
    print(frac[2]);
    printf("\n");

    //product
    sum=numerator[0]*numerator[1];
    res=denominator[0]*denominator[1];
    gcd2=abs(GCD(sum,res));
    transfer(sum/gcd2,res/gcd2,2);
    print(frac[0]);
    printf(" * ");
    print(frac[1]);
    printf(" = ");
    print(frac[2]);
    printf("\n");

    //quotient
    print(frac[0]);
    printf(" / ");
    print(frac[1]);
    printf(" = ");
    if(numerator[1]==0){
        printf("Inf\n");
    }
    else{
        sum=numerator[0]*denominator[1];
        res=denominator[0]*numerator[1];
        if(res<0){
            sum=-sum;
            res=-res;
        }

        gcd2=abs(GCD(sum,res));
        transfer(sum/gcd2,res/gcd2,2);

        print(frac[2]);
        printf("\n");
    }

    return 0;
}