题目地址

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

#include <iostream>
#include <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=40010,inf=100000000;
//struct node{
//    ll numerator,denominator;
//}a[maxn];
ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b){
    ll t=gcd(a,b);
    return a/t*b;
}
int n;ll deno=1,numer=0,ans0,ans1=0,ans2=1,a,b;
int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        scanf("%lld/%lld",&a,&b);
        ans1=a*ans2+b*ans1;
        ans2=b*ans2;
        ll gcd_sum=gcd(ans1,ans2);
        ans1/=gcd_sum;ans2/=gcd_sum;
    }
    if(ans1>ans2){
        ans0=ans1/ans2;
        ans1=ans1%ans2;
    }
    if(ans2==1){
        ans0+=ans1;
        cout<<ans0;return 0;
    }
    if(ans0)
    cout<<ans0<<" ";cout<<ans1<<"/"<<ans2;
}