注意，最后统计答案的时候，要把f[1]也考虑进去，也可以直接把f[0]设为1，就能保证正确性了。
#include <iostream> #include <stdio.h> #include <algorithm> #include <cstring> #define For(i, j, n) for (int i = j; i <= n; ++i) using namespace std; typedef long long LL; const int N = 105; int n; LL a[N]; LL gcd(LL x, LL y) { if (y) return gcd(y, x % y); return x; } LL lcd(LL x, LL y) { return x * y / gcd(x, y); } struct fraction { LL father, son; bool operator>(const fraction &t) const { return son * t.father > t.son * father; } bool operator<(const fraction &t) const { return son * t.father < t.son * father; } fraction operator/(const fraction &t) const { fraction c; c.father = father / t.father; c.son = son / t.son; return c; } bool operator!=(const fraction &t) const { return (father != t.father || son != t.son); } bool operator==(const fraction &t) const { return !(father != t.father || son != t.son); } } f[N]; ostream& operator<<(std::ostream& os, const fraction& p) { os << p.son << "/" << p.father; return os; } int cnt; int main() { cin >> n; for (int i = 1; i <= n; i++) cin >> a[i]; sort(a + 1, a + n + 1); fraction ans; ans.father = 1; ans.son = 1e12; fraction mans; for (int i = 1; i < n; i++) { if(a[i] == a[i + 1]) continue; fraction t; t.son = a[i + 1]; t.father = a[i]; LL g = gcd(a[i], a[i + 1]); t.father /= g; t.son /= g; f[++cnt] = t; } sort(f + 1, f + cnt + 1); for(int i = 1; i < cnt; i++) { if(f[i] == f[i + 1]) continue; fraction t = f[i + 1] / f[i]; ans = min(ans, t); } cout << min(ans, f[1]); return 0; }