PAT(Basic Level) Practice : 1034 有理数四则运算 (20分)

1034 有理数四则运算 (20分)

测试点2：结果错误

“分子和分母全是整形范围内的整数”，但是两个整数相乘会超出整数int的范围。
在未化简之前，乘法过程中，分子分母相乘可能会超出范围。
使用long long ，而不是long（和int一样是4个字节）
用法："%lld"

测试点3：运行超时

运行超时，最坑的地方...
cout转printf，不行。
后来查了一下，要使用辗转相除法求最大公约数，然后来约分才不会超时。

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

gcd返回值有可能是0，要判断一下，不要除以0导致浮点错误

代码

#include <iostream>
#include"stdlib.h"
#include <vector>
#include <string>
#include <cstdio>
//scanf printf防止超时

#include <algorithm>
//vector的sort
#include <sstream>
//转换

using namespace std;

#include<iomanip>
//精度

#include<cmath>
//round四舍五入取整

class number
{
public:
    bool isMinus;
    long long num1;//分子
    long long num2;//分母

    long long integ;//整数部分
    number()
    {
        isMinus=false;
        num1=0;num2=0;integ=0;
    }
};

//辗转相除法求最大公约数
long long gcd(long long a,long long b)
{
    return b==0?a:gcd(b,a%b);
}
//消除公因子
void simplify(number & num)
{
    
    long long temp=gcd(num.num1,num.num2);
    if(temp!=0)
    {
        num.num1/=temp;
        num.num2/=temp;
    }
    

}

//取整数
void simplify1(number & num)
{
    if(num.num1==0)
    {
        num.num2==0;
    }else{
        num.integ=num.num1/num.num2;
        num.num1=num.num1%num.num2;
    }
}
number add(const number & n1,const number & n2)
{
    if(n1.num1==0)
        return n2;
    if(n2.num1==0)
        return n1;

    number res;
    long long  t1,t2;
    if(n1.isMinus)
    {
        t1=(-1)*n1.num1*n2.num2;
    }else{
        t1=n1.num1*n2.num2;
    }

    if(n2.isMinus)
    {
        t2=(-1)*n2.num1*n1.num2;
    }else{
        t2=n2.num1*n1.num2;
    }

    t1+=t2;

    if(t1<0)
    {
        res.isMinus=true;
        res.num1=abs(t1);
    }else{
        res.isMinus=false;
        res.num1=t1;
    }

    res.num2=n1.num2*n2.num2;

    return res;
}

number multiply(const number & n1,const number & n2)
{
    number res;
    if(n1.num1==0||n2.num1==0)
    {
        res.isMinus=false;
        res.num1=0;
        res.num2=0;
        return res;
    }

    if((n1.isMinus&&(!n2.isMinus))||(n2.isMinus&&(!n1.isMinus)))
    {
        res.isMinus=true;
    }else{
        res.isMinus=false;
    }
    res.num1=n1.num1*n2.num1;
    res.num2=n1.num2*n2.num2;
    return res;

}


void output(number n)
{
    if(n.integ==0&&n.num1==0)
    {
        printf("0");
    }else if(n.isMinus){
        //cout<<"(-";
        printf("(-");
        if(n.integ==0)
        {
            //cout<<n.num1<<"/"<<n.num2;
            printf("%lld/%lld",n.num1,n.num2);
        }else if(n.num1==0)
        {
            //cout<<n.integ;
            printf("%lld",n.integ);
        }else{
            //cout<<n.integ<<" "<<n.num1<<"/"<<n.num2;
            printf("%lld %lld/%lld",n.integ,n.num1,n.num2);
        }

        //cout<<")";
        printf(")");
    }else{
        if(n.integ==0)
        {
            //cout<<n.num1<<"/"<<n.num2;
            printf("%lld/%lld",n.num1,n.num2);
        }else if(n.num1==0)
        {
            //cout<<n.integ;
            printf("%lld",n.integ);
        }else{
            //cout<<n.integ<<" "<<n.num1<<"/"<<n.num2;
            printf("%lld %lld/%lld",n.integ,n.num1,n.num2);
        }
    }
}
int main()
{
    long long  a1,b1,a2,b2;
    scanf("%lld/%lld %lld/%lld",&a1,&b1,&a2,&b2);

    number n1,n2;
    n1.isMinus=a1<0;
    n1.num1=abs(a1);
    n1.num2=b1;

    n2.isMinus=a2<0;
    n2.num1=abs(a2);
    n2.num2=b2;

    simplify(n1);
    simplify(n2);


   
    //+
    number res1;
    res1=add(n1,n2);
    simplify(res1);

    number n1_out=n1;
    number n2_out=n2;
    simplify1(n1_out);
    simplify1(n2_out);
    simplify1(res1);
    output(n1_out);
    //cout<<" + ";
    printf(" + ");
    output(n2_out);
    //cout<<" = ";
    printf(" = ");
    output(res1);
    //cout<<endl;
    printf("\n");

    //-
    number n3;
    if(n2.isMinus)
    {
        n3.isMinus=false;
    }else{
        n3.isMinus=true;
    }
    n3.num1=n2.num1;
    n3.num2=n2.num2;

    number res2;
    res2=add(n1,n3);
    simplify(res2);

    simplify1(res2);
    output(n1_out);
    //cout<<" - ";
    printf(" - ");
    output(n2_out);
    //cout<<" = ";
    printf(" = ");
    output(res2);
    //cout<<endl;
    printf("\n");

    //*
    number res3;
    res3=multiply(n1,n2);
    simplify(res3);
    simplify1(res3);
    output(n1_out);
    //cout<<" * ";
    printf(" * ");
    output(n2_out);
    //cout<<" = ";
    printf(" = ");
    output(res3);
    //cout<<endl;
    printf("\n");


    //division
    output(n1_out);
    //cout<<" / ";
    printf(" / ");
    output(n2_out);
    //cout<<" = ";
    printf(" = ");
    if(n2.num1==0)
    {
        //cout<<"Inf"<<endl;
        printf("Inf\n");
    }
    else{
        int temp=n2.num1;
        n2.num1=n2.num2;
        n2.num2=temp;
        number res4=multiply(n1,n2);
        simplify(res4);
        simplify1(res4);
        output(res4);
        //cout<<endl;
        printf("\n");
    }
    return 0;
}