【PAT】A1081 Rational Sum (20分)（分数的四则运算）

1081 Rational Sum (20分)

题目链接

代码：

#include<iostream>
#include<cmath>
using namespace std;

struct Fraction{
	int up;
	int down;
	Fraction()
	{
		up = 0;
		down= 1;//注意这里是1，而不是0
	}
}*fraction;

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

//如果分母为负数，则分子分母同时取相反数；
//如果分子为0，则分母为1；
//如果分子分母有公约数，则化简 
Fraction reduction(Fraction a)
{
	if(a.down < 0){
		a.down = -a.down;
		a.up = -a.up;
	}
	else if(a.up == 0)
		a.down = 1;
	else{
		int d = gcd(a.up, a.down);
		a.up /= d;
		a.down /= d;
	}
	return a;
}

Fraction add(Fraction a, Fraction b)
{
	Fraction temp;
	temp.up = a.up * b.down + a.down * b.up;
	temp.down = a.down * b.down;
	return reduction(temp);
}

void print(Fraction a)
{
	if(a.down == 1)
		cout<<a.up<<endl;
	else if(a.up == 0)
		cout<<"0"<<endl;
	else if(abs(a.up) > a.down)
		cout<<a.up/a.down<<" "<<abs(a.up) % a.down<<"/"<<a.down<<endl;
	else{
		cout<<a.up<<"/"<<a.down<<endl;
	}
}

int main()
{
	int n;
	cin>>n;
	fraction = new Fraction[n+5];
	for(int i=0;i<n;i++)
	{
		scanf("%d/%d", &fraction[i].up, &fraction[i].down);
	}
	Fraction sum;
	for(int i=0;i<n;i++){
		sum = add(sum, fraction[i]);
	}
	print(sum);
	delete []fraction;
	return 0;
}

其实，main函数还可以简化：

int main()
{
	int n;
	cin>>n;
	fraction = new Fraction[n+5];
	Fraction sum;
	for(int i=0;i<n;i++)
	{
		scanf("%d/%d", &fraction[i].up, &fraction[i].down);
		sum = add(sum, fraction[i]);
	}
	print(sum);
	delete []fraction;
	return 0;
}