PAT basic 1053

#include <iostream>using namespace std;int main() {    int n, d, k, maybe = 0, must = 0;    double e, temp;    cin >> n >> e >> d;    for (int i = 0; i < n; i++) {        cin >> k;        int sum = 0;        for (int j = 0; j < k; j++) {            cin >> temp;            if (temp < e) sum++;        }        if(sum > (k / 2)) {            k > d ? must++ : maybe++;        }    }    double mayberesult = (double)maybe / n * 100;    double mustresult = (double)must / n * 100;    printf("%.1f%% %.1f%%", mayberesult, mustresult);    return 0;}