PAT_1088. Rational Arithmetic

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/kukubao207/article/details/77575517

#include <iostream>
using namespace std;
long long int gcd(long long int m,long long int n)
{
	if(n==0)
		return m;
	else
		return gcd(n,m%n);
}
void outputStandard(long long int a,long long int b)
{
	bool isNegative=false;
	if(a<0&&b>0||(a>0&&b<0))
	{
		cout<<"(-";
		isNegative=true;
		a=abs(a);
		b=abs(b);
	}
	long long int G=gcd(a,b);
	a/=G;
	b/=G;

	if(b==1)
		cout<<a;
	else if(a>b)
		cout<<a/b<<" "<<a%b<<"/"<<b;
	else
		cout<<a%b<<"/"<<b;
	if(isNegative)
		cout<<")";
}
void compute(long long int a1,long long int b1,long long int a2,long long int b2,char sign)
{
	//求绝对值
	bool isNegative1=false,isNegative2=false;
	if(a1<0)
	{
		a1=-a1;
		isNegative1=true;
	}
	if(a2<0)
	{
		a2=-a2;
		isNegative2=true;
	}

	//约分
	long long int G=gcd(a1,b1);
	a1/=G;
	b1/=G;
	G=gcd(a2,b2);
	a2/=G;
	b2/=G;

	if(isNegative1==true)
		a1=-a1;
	if(isNegative2==true)
		a2=-a2;

	if(sign=='+')
	{
		a1=a1*b2+a2*b1;
		b1=b1*b2;
	}
	else if(sign=='-')
	{
		a1=a1*b2-a2*b1;
		b1=b1*b2;
	}
	else if(sign=='*')
	{
		a1=a1*a2;
		b1=b1*b2;
	}
	else if(sign=='/')
	{
		if(a2==0)
		{
			cout<<"Inf";
			return ;
		}
		a1=a1*b2;
		b1=b1*a2;
	}
	outputStandard(a1,b1);
}
int main()
{
	long long int a1,b1,a2,b2;
	scanf("%lld/%lld %lld/%lld",&a1,&b1,&a2,&b2);

	outputStandard(a1,b1);
	cout<<" + ";
	outputStandard(a2,b2);
	cout<<" = ";
	compute(a1,b1,a2,b2,'+');
	cout<<endl;

	outputStandard(a1,b1);
	cout<<" - ";
	outputStandard(a2,b2);
	cout<<" = ";
	compute(a1,b1,a2,b2,'-');
	cout<<endl;

	outputStandard(a1,b1);
	cout<<" * ";
	outputStandard(a2,b2);
	cout<<" = ";
	compute(a1,b1,a2,b2,'*');
	cout<<endl;

	outputStandard(a1,b1);
	cout<<" / ";
	outputStandard(a2,b2);
	cout<<" = ";
	compute(a1,b1,a2,b2,'/');
	cout<<endl;
	return 0;
}

这题与PAT1081 类似，都是要用到辗转相除法求最大公约数的知识点的。

两题的思路也是类似，在进行分数的运算之前，先对分数进行约分，之后再进行运算。