L1-009 N个数求和
#include <iostream>
#include <algorithm>using namespace std;typedef long long ll;
const int N = 105;typedef struct node {ll x, y;
}node;
node a[N];ll gcd(ll a, ll b)
{return b ? gcd(b, a % b) : a;
}int main()
{int n;cin >> n;for (int i = 0; i < n; i++) scanf("%lld/%lld", &a[i].x, &a[i].y);ll ans = a[0].y;for (int i = 1; i < n; i++){ans = ans / gcd(ans, a[i].y) * a[i].y;}//printf("%lld\n", ans);ll res = 0; //分子和for (int i = 0; i < n; i++){res = res + ans / a[i].y * a[i].x;}ll x = res / ans;ll b = res % ans;ll c = 1;if (b != 0){c = gcd(b, ans);b /= c, ans /= c;}if (x == 0){if (b != 0)printf("%lld/%lld", b, ans);else printf("0");}if (b == 0 && x != 0) printf("%lld", x);if(x!=0 && b!= 0) printf("%lld %lld/%lld", x, b, ans);return 0;
}
