题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805386161274880
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：题意 给出n个分数，求分数的和，并化为分数的形式输出。
#include<iostream>
using namespace std;
typedef long long ll;
ll gcd(int x,int y){
if(!y){
return x;
}else{
return gcd(y,x%y);
}
}
int main(){
ll n,a,b;
ll sum_a=0,sum_b=1;
ll val;
scanf("%lld",&n);
for(int i=0;i<n;i++){
scanf("%lld/%lld",&a,&b);
val=gcd(a,b);
a=a/val;
b=b/val;
sum_a=a*sum_b+b*sum_a;
sum_b=b*sum_b;
val=gcd(sum_a,sum_b);
sum_a=sum_a/val;
sum_b=sum_b/val;
}
if(sum_a==0){
printf("0\n");
}else{
ll ans=sum_a/sum_b;
sum_a=sum_a-(sum_b*ans);
if(ans!=0){
if(sum_a!=0){
printf("%lld %lld/%lld",ans,sum_a,sum_b);
}else{
printf("%lld",ans);
}
}else{
printf("%lld/%lld",sum_a,sum_b);
}
}
return 0;
}