PAT 1081. Rational Sum

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

分析

用辗转相除法求最大公因数，否则超时

#include<iostream>
#include<math.h>
using namespace std;
long long int n,s1,s2,m1,m2;
long long int gcd(long long int a,long long int b){
	if(a==0) return 1;
    return a%b==0?b:gcd(b,a%b);
}
void add(){
	s1=s1*m2+s2*m1;
	m1=m1*m2;
	long long int s=gcd(abs(s1),m1);
	m1/=s; s1/=s;
}
int main(){	
    cin>>n;
	scanf("%lld/%lld",&s1,&m1);
	for(int i=1;i<n;i++){
		scanf("%lld/%lld",&s2,&m2);
		add();
	}
	if(s1<0) cout<<"-";
	long long int t=abs(s1)/m1;
	s1=abs(s1)%m1; 
	if(s1==0) printf("%lld",t);// 整数 
	else if(t!=0)
	    printf("%lld %lld/%lld",t,s1,m1); //假分数 
	else
	    printf("%lld/%lld",s1,m1); //分数 
        return 0;
}