题目传送门
#include <iostream> using namespace std; int main() { int n, d, k, may = 0, must = 0; double e, t; cin >> n >> e >> d; for (int i = 0; i < n; i++) { cin >> k; int sum = 0; for (int j = 0; j < k; j++) { cin >> t; if (t < e) { sum++; } } if (sum > (k / 2)) { k > d ? must++ : may++; } } double x = (double)may / n * 100; double y = (double)must / n * 100; printf("%.1f%% %.1f%%", x, y); return 0; }