#include<iostream>
using namespace std;
typedef long long LL;
LL gcd(LL a, LL b)
{
return b ? gcd(b, a % b) : a;
}
int main()
{
LL N = 0, a = 0, b = 0, sum_a = 0, sum_b = 1, t;
scanf("%lld", &N);
for (int i = 0; i < N; i++)
{
scanf("%lld/%lld", &a, &b);
//通分
sum_a *= b;
sum_a += a * sum_b;
sum_b *= b;
//约分
t = gcd(sum_a, sum_b);
sum_a /= t;
sum_b /= t;
}
if (sum_a % sum_b == 0)
printf("%lld\n", sum_a / sum_b);
else if (sum_a < sum_b)
printf("%lld/%lld\n", sum_a, sum_b);
else
printf("%lld %lld/%lld\n", sum_a / sum_b, sum_a % sum_b, sum_b);
return 0;
}
天梯练习 7-9 N个数求和
最新推荐文章于 2025-03-16 17:18:06 发布