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

最新推荐文章于 2021-12-18 14:48:30 发布

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最新推荐文章于 2021-12-18 14:48:30 发布

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本文介绍了一个程序设计案例，用于实现两个有理数的基本算术运算，包括加法、减法、乘法和除法。通过解析输入字符串提取分子和分母，并进行约分处理，最终输出最简形式的结果。

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1088. Rational Arithmetic (20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient.

Input Specification:

Each input file contains one test case, which gives in one line the two rational numbers in the format "a1/b1 a2/b2". The numerators and the denominators are all in the range of long int. If there is a negative sign, it must appear only in front of the numerator. The denominators are guaranteed to be non-zero numbers.

Output Specification:

For each test case, print in 4 lines the sum, difference, product and quotient of the two rational numbers, respectively. The format of each line is "number1 operator number2 = result". Notice that all the rational numbers must be in their simplest form "k a/b", where k is the integer part, and a/b is the simplest fraction part. If the number is negative, it must be included in a pair of parentheses. If the denominator in the division is zero, output "Inf" as the result. It is guaranteed that all the output integers are in the range of long int.

Sample Input 1:

2/3 -4/2

Sample Output 1:

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

Sample Input 2:

5/3 0/6

Sample Output 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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1088

https://www.nowcoder.com/pat/5/problem/4036

思路：

主要是数字的提取

注意各种特殊情况

注意数据范围，稳妥起见全部使用longlong

CODE：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cstdlib>

using namespace std;

typedef struct  S
{
	long long fz;
	long long fm;	
};


long long gcd(long long a,long long b)
{
	if ( max(a,b)%min(a,b)==0)
	{
		return min(a,b);
	}
	else
	{
		return gcd(min(a,b),max(a,b)-min(a,b));
	}
}


long long lcm(long long a,long long b)
{
	return a/gcd(abs(a),abs(b))*b;
} 

S pre(string a)
{
	S n;
	long long fh;
	for (long long i=0;i<a.length();i++) 
		if (a[i]=='/')
			{
				fh=i;
				break;
			} 
	if (a[0]=='-')
	{
		n.fz=-1*atoi(a.substr(1,fh-1).c_str());
	}
	else
		n.fz=atoi(a.substr(0,fh).c_str());
	n.fm=atoi(a.substr(fh+1,a.length()-fh-1).c_str());
	if (n.fz!=0)
	{
		long long gc=gcd(abs(n.fz),n.fm);
		n.fz/=gc;
		n.fm/=gc;
	}
	return n;
}
void prt(S a)
{
	if (a.fz==0)
	{
		cout<<"0";
	}
	else
	{
		if (a.fz<0)
		{
			cout<<"(-";
			a.fz*=-1;
			if (a.fz>=a.fm)
			{
				cout<<a.fz/a.fm;
				a.fz%=a.fm;
				if (a.fz!=0)
				{	
					cout<<" "<<a.fz<<"/"<<a.fm;
				}
			}
			else
				cout<<a.fz<<"/"<<a.fm;
			cout<<")";
		}
		else
		{
			if (a.fz>=a.fm)
			{
				cout<<a.fz/a.fm;
				a.fz%=a.fm;
				if (a.fz!=0)
				{	
					cout<<" "<<a.fz<<"/"<<a.fm;
				}
			}
			else
				cout<<a.fz<<"/"<<a.fm;
		}
	}	
}
void opr(S x,S y,char fh)
{
	
	prt(x);
	cout<<" "<<fh<<" ";
	prt(y);
	cout<<" = ";
	long long nfm=lcm(x.fm,y.fm);
	long long fz1=x.fz*(nfm/x.fm);
	long long fz2=y.fz*(nfm/y.fm);
	//cout<<nfm<<" "<<fz1<<" "<<fz2<<endl; 
	S ad;
	if (fh=='+')
	{
		ad.fz=fz1+fz2;
		ad.fm=nfm;
	}
	if (fh=='-')
	{
		ad.fz=fz1-fz2;
		ad.fm=nfm;
	}
	if (fh=='*')
	{
		ad.fz=x.fz*y.fz;
		ad.fm=x.fm*y.fm;
	}
	if (fh=='/')
	{
		if (y.fz==0)
		{
			cout<<"Inf";
			return;
		}
		ad.fz=x.fz*y.fm;
		ad.fm=x.fm*y.fz;
		if (ad.fm<0)
		{
			ad.fm*=-1;
			ad.fz*=-1;
		}
	}
	if (ad.fz!=0)
	{
		long long sgc=gcd(abs(ad.fz),ad.fm);
		ad.fz/=sgc;
		ad.fm/=sgc;
		prt(ad);
		cout<<endl;
	}
	else
		cout<<"0"<<endl;	
}
int main()
{
	S x,y;
	string a,b;
	cin>>a>>b;
	x=pre(a);
	y=pre(b);
	//cout<<x.fz<<" "<<x.fm<<endl;
	//cout<<y.fz<<" "<<y.fm<<endl;
	opr(x,y,'+');
	opr(x,y,'-');
	opr(x,y,'*');
	opr(x,y,'/');
	return 0;
}