PAT A1081：Rational Sum之分数加法

题目描述

1081 Rational Sum (20分)
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:

2

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

求解思路

用一个结构体关联分数的分子和分母，进行分数的相加与化简运算即可。

代码实现(AC)

#include<cstdio>
#include<cmath>
struct Fraction{
	long long up,down;
}; 
int gcd(int a,int b)
{
	return !b?a:gcd(b,a%b);
}
/*分数化简*/
Fraction reduction(Fraction result)
{
	if(result.down<0)	//分母为负数，令分子和分母都变为相反数 
	{
		result.up=(-result.up);
		result.down=(-result.down);
	}
	if(result.up==0)	//如果分子为0，则令分母为1 
	{
		result.down=1;
	}
	else	//如果分子不为零，则进行约分，约去最大公约数 
	{
		int d=gcd(abs(result.up),abs(result.down));
		result.up/=d;
		result.down/=d;
	} 
	return result; 
}
Fraction add(Fraction f1,Fraction f2)
{
	Fraction result;
	result.up=f1.up*f2.down+f1.down*f2.up;
	result.down=f1.down*f2.down;
	return reduction(result);	//结果注意化简 
}

void showResult(Fraction r)
{
	reduction(r);
	if(r.down==1)	printf("%lld\n",r.up);	//整数
	else if(abs(r.up)>r.down)
	{
		printf("%lld %lld/%lld\n",r.up/r.down,abs(r.up)%r.down,r.down);	//假分数	
	} 
	else
	{
		printf("%lld/%lld",r.up,r.down);	//真分数 
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	Fraction sum,temp;
	sum.up=0;
	sum.down=1;
	for(int i=0;i<n;i++)
	{
		scanf("%lld/%lld\n",&temp.up,&temp.down);
		sum=add(sum,temp);
	}
	showResult(sum);
	return 0;
}