PAT (Advanced Level) 1081. Rational Sum (20) 分數求和，輾轉相除法求gcd

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

輾轉相除法求最大公約數。

/*2015.7.30cyq*/
#include <iostream>
#include <string>
using namespace std;

long gcd(long a,long b){//輾轉相除求最大公約數
	long r;
	while(b!=0){
		r=a%b;
		a=b;
		b=r;
	}
	return a;
}

void add(long &a,long &b,long &a1,long &b1){
	long flag=1;
	long num=a*b1+a1*b;
	long den=b*b1;
	if(num<0){
		flag=-1;
		num=-num;
	}
	long g=gcd(num,den);
	a=flag*num/g;
	b=den/g;
}

int main(){
	int N;
	cin>>N;
	long a1,b1,a,b;
	scanf("%ld/%ld",&a,&b);
	for(int i=1;i<N;i++){
		scanf("%ld/%ld",&a1,&b1);
		add(a,b,a1,b1);
	}
	if(a==0){
		printf("0");
		return 0;
	}
	if(b==1){
		printf("%ld",a);
		return 0;
	}
	if(a/b){//整數部分
		printf("%ld ",a/b);
		a=a%b;
	}
	printf("%ld/%ld\n",a,b);
	
	return 0;
}