/*
# 问题分析
题设要求统计符合住房空置条件的房屋比例。这是一道统计题。
需要统计的数据是, 可能空置的房屋数量, 空置的房屋数量;
可能空置的条件: 比较每天的用电量是否小于给定用电量，统计满足条件的天数, 如果天数超过一半, 则为可能空置.
空置的条件: 对于已经认定是可能空置的房间, 比较用电量的天数大于给定天数， 如果大于，则认为是空置。
 
# 完整步骤描述
1. 获取输入: 房屋数量, 最低可能空置用电量, 最低空置观察天数
2. 初始化可能空置的房屋计数器, 空置的房屋计数器
3. 对于每一个房屋:
    - 获取观察天数
    - 初始化用电量低天数的计数器为0
    - 对于每一天的用电量:
        - 如果用电量小于阈值, 则天数计数器+1
    - 如果天数计数器的值的2倍大于观察天数且观察天数大于指定天数, 则空置的房屋计数器+1
    - 否则, 如果天数计数器的值的2倍大于观察天数, 则可能空置的房屋计数器+1
4. 按格式输出可能空置和空置的比例
 
# 伪代码描述
1. get input: house_amount, min_electricity, min_days
2. init counter:
    - maybe_empty_house_amount
    - empty_house_amount
3. for each house:
    - get input: observation_days and electricity of each day
    - init counter: low_electricity_days
    - for each electricity:
        - if electricity < min_electricity: low_electricity_days++
    - if low_electricity_days * 2 > observation_days and observation_days > min_days:
        - empty_house_amount++
    - else if low_electricity_days * 2 > observation_days:
        - maybe_empty_house_amount++
 
4. calculate rate:
    - maybe_empty_house_rate = maybe_empty_house_amount / house_amount * 100
    - empty_house_rate = empty_house_amount / house_amount * 100
5. print(maybe_empty_house_rate, empty_house_rate)
 
*/
 
 
#include <stdio.h>
#define EPS 1e-6
 
int main() {
  int house_amount;
  double min_electricity;
  int min_days;
  scanf("%d %lf %d", &house_amount, &min_electricity, &min_days);
 
  int maybe_empty_house_amount = 0;
  int empty_house_amount = 0;
  int observation_days;
  double daily_electricity;
  for (int i = 0; i < house_amount; i++) {
    scanf("%d", &observation_days);
 
    int low_electricity_days = 0;
    for (int j = 0; j < observation_days; j++) {
      scanf("%lf", &daily_electricity);
      if (daily_electricity + EPS < min_electricity) {
        low_electricity_days++;
      }
    }
 
    if (low_electricity_days * 2 > observation_days && observation_days > min_days) {
      empty_house_amount++;
    } else if (low_electricity_days * 2 > observation_days) {
      maybe_empty_house_amount++;
    }
  }
 
  double maybe_empty_house_rate = (double)maybe_empty_house_amount / house_amount * 100;
  double empty_house_rate = (double)empty_house_amount / house_amount * 100;
 
  printf("%.1f%% %.1f%%", maybe_empty_house_rate, empty_house_rate);
 
  return 0;
}