A1081 Rational Sum

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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.

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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

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#include<bits/stdc++.h> 
#define MAX 110
using namespace std;
struct fract{
	int a,b;

};
int n;
int gcd(int a,int b){
	if(a%b==0) return b;
	gcd(b,a%b);
}
fract f[MAX];
fract redu(fract m){
	fract res;
	if(m.b<0){
		m.b=-m.b;
		m.a=-m.a;
	} 
	int g=gcd(abs(m.a),abs(m.b));
	res.a=m.a/g;
	res.b=m.b/g;
	return res;
}
fract add(fract m,fract n){
	fract res;
	res.b=m.b*n.b;
	res.a=m.a*n.b+n.a*m.b;
	return redu(res); 
	 
	
}
void show(fract ff){
	if(ff.b==1) {
	cout<<ff.a<<endl;
return;

}
else if(abs(ff.a)>ff.b){
	int t=ff.a/ff.b;
	int wei=ff.a-ff.b*t;
	cout<<t<<" "<<wei<<"/"<<ff.b<<endl;
}
else cout<<ff.a<<"/"<<ff.b<<endl;
}
int main(){
	fract sum;
	sum.a=0;
	sum.b=1;
	cin>>n;
	for(int i=0;i<n;i++){
		scanf("%d/%d",&f[i].a,&f[i].b);
	}
	for(int i=0;i<n;i++){
		sum=add(sum,f[i]);
	}
	show(sum);
	
}