Problem:
Score : 400 points

Problem Statement

You are given N items.

The value of the i-th item (1 \leq i \leq N) is v_i.

Your have to select at least A and at most B of these items.

Under this condition, find the maximum possible arithmetic mean of the values of selected items.

Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

1 \leq N \leq 50

1 \leq A,B \leq N

1 \leq v_i \leq 10^{15}

Each v_i is an integer.

Input

The input is given from Standard Input in the following format:

N A B
v_1
v_2
...
v_N

Output

Print two lines.

The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.

The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Sample Input 1

5 2 2
1 2 3 4 5

Sample Output 1

4.500000
1

The mean of the values of selected items will be maximized when selecting the fourth and fifth items. Hence, the first line of the output should contain 4.5.

There is no other way to select items so that the mean of the values will be 4.5, and thus the second line of the output should contain 1.

Sample Input 2

4 2 3
10 20 10 10

Sample Output 2

15.000000
3

There can be multiple ways to select items so that the mean of the values will be maximized.

Sample Input 3

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996

Sample Output 3

1000000000000000.000000
1

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h> // int64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int(void) {
  int num;
  scanf("%d", &num);
  return num;
}

int get_int2(int *a1, int *a2) {
  scanf("%d %d", a1, a2);
  return 0;
}

// <arr[0]> <arr[1]> .. <arr[size-1]>
int fget_array(int64_t *arr, int size) {
    int i;
    for(i = 0; i < size; i++) {
        scanf("%lld", &arr[i]);
    }
    return 0;
}

// sort
int desc(const void *a1, const void *a2) {
    int64_t aa1 = *(int64_t*)a1;
    int64_t aa2 = *(int64_t*)a2;
    if(aa1 > aa2) return -1;
    if(aa1 == aa2) return 0;
    return 1;
}

#define NUM_MAX 50

static int64_t combi[NUM_MAX+1][NUM_MAX+1];
void init_combi(int num){
    for(int i=0; i<=num; i++) {
        for(int j=0; j<=i; j++) {
            if(j == 0 || j == i) combi[i][j] = 1;
            else {
                combi[i][j] = (combi[i-1][j-1] + combi[i-1][j]);
            }
        }
    }
}

int main(void) {
    int num = get_int();
    int ulimit, llimit;
    get_int2(&llimit, &ulimit);
    static int64_t arr[NUM_MAX+1];
    fget_array(arr, num);
    init_combi(NUM_MAX);
    qsort(arr, num, sizeof(int64_t), desc);
    
    int i;
    double sum = 0;
    for(i = 0; i < llimit; i++) {
        int64_t cur = arr[i];
        sum += cur;
    }
    int64_t prev = arr[llimit-1];
    double ave = sum / llimit;

    int64_t ans = 0;
    if(prev == arr[0]) {
        for(i = llimit; i < ulimit; i++) {
            if(prev != arr[i]) break;
        }
        int cnt = i;
        // [llimit-1, i) -> [llimit, i+1)
        for(i = llimit; i <= cnt; i++) {
            ans += combi[cnt][i];
        }
    } else {
        int k = 0;
        int n = 0;
        for(i = num-1; i >= 0; i--) {
            if(prev == arr[i]) {
                n++;
                if(i < llimit) k++;
            }
        }
        ans = combi[n][k];
    }

    printf("%.8lf\n", ave);
    printf("%lld\n", ans);
    return 0;
}