PAT (Advanced Level) 1088. Rational Arithmetic (20)

Posted Fighting Heart

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了PAT (Advanced Level) 1088. Rational Arithmetic (20)相关的知识,希望对你有一定的参考价值。

简单题。

注意:读入的分数可能不是最简的。输出时也需要转换成最简。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;

struct FS
{
    long long fz,fm;
    FS(long long a,long long b)
    {
        fz=a;
        fm=b;
    }
};

long long gcd(long long a,long long b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}

FS change(FS res)
{
    if(res.fz!=0)
    {
        long long GCD=gcd(abs(res.fz),abs(res.fm));
        res.fz=res.fz/GCD;
        res.fm=res.fm/GCD;
    }

    else
    {
        res.fz=0;
        res.fm=1;
    }

    return res;
}

FS ADD(FS a,FS b)
{
    FS res(0,1);

    res.fz=a.fz*b.fm+b.fz*a.fm;
    res.fm=a.fm*b.fm;

    res=change(res);
    return res;
}

FS SUB(FS a,FS b)
{
    FS res(0,1);

    res.fz=a.fz*b.fm-b.fz*a.fm;
    res.fm=a.fm*b.fm;

    res=change(res);
    return res;
}

FS MUL(FS a,FS b)
{
    FS res(0,1);

    res.fz=a.fz*b.fz;
    res.fm=a.fm*b.fm;

    res=change(res);
    return res;
}

FS DIV(FS a,FS b)
{
    FS res(0,1);

    if(b.fz==0)
    {
        res.fz=0;
        res.fm=0;
        return res;
    }

    if(a.fz==0) return res;

    res.fz=a.fz*b.fm;
    res.fm=a.fm*b.fz;

    if(res.fm<0)
    {
        res.fm=-res.fm;
        res.fz=-res.fz;
    }

    res=change(res);
    return res;
}

void output(FS a)
{
    if(a.fm==0)
    {
        printf("Inf");
        return;
    }

    if(a.fz==0)
    {
        printf("0");
        return;
    }

    a=change(a);
    if(abs(a.fz)<a.fm)
    {
        if(a.fz<0) printf("(");

        printf("%lld/%lld",a.fz,a.fm);
        if(a.fz<0) printf(")");
        return;
    }

    if(a.fz>0)
    {
        if(a.fz%a.fm==0)
        {
            printf("%lld",a.fz/a.fm);
            return;
        }
        else
        {
            printf("%lld %lld/%lld",a.fz/a.fm,a.fz%a.fm,a.fm);
            return;
        }
    }

    else
    {
        a.fz=-a.fz;

        printf("(");
        if(a.fz%a.fm==0)
        {
            printf("-%lld",a.fz/a.fm);
            printf(")");
            return;
        }
        else
        {
            printf("-%lld %lld/%lld",a.fz/a.fm,a.fz%a.fm,a.fm);
            printf(")");
            return;
        }

    }
}

int main()
{
    long long s1,s2,s3,s4;
    scanf("%lld/%lld %lld/%lld",&s1,&s2,&s3,&s4);

    FS a(s1,s2);
    FS b(s3,s4);

    output(a); cout<<" + "; output(b); cout<<" = "; output(ADD(a,b));
    cout<<endl;
    output(a); cout<<" - "; output(b); cout<<" = "; output(SUB(a,b));
    cout<<endl;
    output(a); cout<<" * "; output(b); cout<<" = "; output(MUL(a,b));
    cout<<endl;
    output(a); cout<<" / "; output(b); cout<<" = "; output(DIV(a,b));
    cout<<endl;

    return 0;
}

 

以上是关于PAT (Advanced Level) 1088. Rational Arithmetic (20)的主要内容,如果未能解决你的问题,请参考以下文章

PAT (Advanced Level) 1088. Rational Arithmetic (20)

PAT (Advanced Level) 1025. PAT Ranking (25)

PAT Advanced Level 1044

PAT Advanced Level 1043

PAT Advanced Level 1079

PAT Advanced Level 1095