PAT A1081. Rational Sum (20)

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.

52/5 4/15 1/30 -2/60 8/3Sample Output 1:
3 1/3Sample Input 2:
24/3 2/3Sample Output 2:
2Sample Input 3:
31/3 -1/6 1/8Sample Output 3:
7/24
#include <cstdio>#include <algorithm>#include <cstring>#include <string>#define Max 111using namespace std;struct Sum{	long long numerator,denominator;}S[Max];long long gcd(Sum a){	long long c,d,e;	d=a.numerator;	e=a.denominator;	if(d<0) d=-d;	if(e<0) e=-e;	if(e>d) swap(d,e);	c=e;	while(d%e!=0)	{		c=d%e;		d=e;		e=c;	}	return c;}Sum MikEase( Sum a){	if(a.denominator<0)	{		a.denominator=-a.denominator;		a.numerator=-a.numerator;	}	if(a.numerator==0) a.denominator=1;	else {	long long k=gcd(a);	a.numerator=a.numerator/k;	a.denominator=a.denominator/k;	}	return a;}Sum Get(Sum a ,Sum b){	a.numerator=a.numerator*b.denominator+b.numerator*a.denominator;	a.denominator=a.denominator*b.denominator;	return MikEase(a);}int main(){	int n;	scanf("%d",&n);	for(int i=0;i<n;i++)	{		scanf("%lld/%lld",&S[i].numerator,&S[i].denominator);		if(i>0)		S[i]=Get(S[i],S[i-1]);	}		long long a,b,c=0;	a=S[n-1].numerator;b=S[n-1].denominator;	if(b==1) PRintf("%lld",a);	else 	{		if(abs(a)>abs(b))		{			printf("%lld %lld/%lld",a/b,abs(a)%abs(b),b);		}		else printf("%lld/%lld",a,b);	}	system("pause");	return 0;}