PAT A 1081. Rational Sum (20)

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本文介绍了一种用于计算多个分数之和的算法，并提供了一个具体的C++实现案例。该算法能够处理正负分数，并将结果简化为最简形式。

题目

Given N rational numbers in the form "numerator/denominator", you are supposed to calculate their sum.

Input Specification:

Each input file contains one test case. Each case starts with a positive integer N (<=100), followed in the next line N rational numbers "a1/b1 a2/b2 ..." where all the numerators and denominators are in the range of "long int". If there is a negative number, then the sign must appear in front of the numerator.

Output Specification:

For each test case, output the sum in the simplest form "integer numerator/denominator" where "integer" is the integer part of the sum, "numerator" < "denominator", and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.

Sample Input 1:

5
2/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

2
4/3 2/3

Sample Output 2:

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

分数求和。

不能先转成小数，必须保持分数格式计算。

直接模拟，注意和为0的特殊情况。

代码：

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

long long com_mul(long long n1,long long n2);	//求最小公倍数
long long com_factor(long long n1,long long n2);	//求最大公因数

int main()
{
	int n;
	cin>>n;
	long long a1=0,b1=1,a2,b2;
	long long mul,factor;	//最小公倍数，最大公因数
	for(int i=0;i<n;i++)
	{
		cin>>a2;	//输入数据
		cin.get();
		cin>>b2;

		mul=com_mul(b1,b2);	//求和
		a1*=mul/b1;
		a2*=mul/b2;
		a1+=a2;
		b1=mul;

		factor=com_factor(a1,b1);	//约简
		a1/=factor;
		b1/=factor;
		if(a1==0)
			b1=1;
	}

	if(a1==0)	//输出
		cout<<0;
	else
	{
		if(a1/b1!=0)
		{
			cout<<a1/b1;
			a1%=b1;
			if(a1!=0)
				cout<<" ";
		}
		if(a1!=0)
		{
			cout<<a1<<"/"<<b1;
		}
	}

	return 0;
}

long long com_mul(long long n1,long long n2)
{
	long long ans=n1*n2;
	long long i=2,limit=sqrt((double)min(n1,n2))+1;
	while(i<=n1&&i<=n2&&i<=limit)
	{
		if(n1%i==0&&n2%i==0)
		{
			ans/=i;
			n1/=i;
			n2/=i;
		}
		else
			i++;
	}
	return ans;
}

long long com_factor(long long n1,long long n2)
{
	long long ans=1;
	long long i=2,limit=sqrt((double)min(n1,n2))+1;
	while(i<=n1&&i<=n2&&i<=limit)
	{
		if(n1%i==0&&n2%i==0)
		{
			ans*=i;
			n1/=i;
			n2/=i;
		}
		else 
			i++;
	}
	return ans;
}