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++14 (GCC 5.4.1) (judge verdict: Accepted)

/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
//const ll mod = 998244353;


inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }

#define si(x) scanf("%d",&x)
#define sii(x,y) scanf("%d %d",&x,&y)
#define siii(x,y,z) scanf("%d %d %d",&x,&y,&z)
#define sl(x) scanf("%lld",&x)
#define sll(x,y) scanf("%lld %lld",&x,&y)
#define slll(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)
#define ss(ch) scanf("%s",ch)
#define pi(x) printf("%d",x)
#define pii(x,y) printf("%d %d",x,y)
#define piii(x,y,z) printf("%d %d %d",x,y,z)
#define pl(x) printf("%lld",x)
#define pll(x,y) printf("%lld %lld",x,y)
#define plll(x,y,z) printf("%lld %lld %lld",x,y,z)
#define ps(ch) printf("%s",ch)
#define F(i,a,b)      for(int i= a; i <= b; i++)
#define R(i,b,a)      for(int i= b; i >= a; i--)

/* for Random Number generate
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
*/
///**
template < typename F, typename S >ostream& operator << ( ostream& os, const pair< F, S > & p ) {return os << "(" << p.first << ", " << p.second << ")";}
template < typename T >ostream &operator << ( ostream & os, const vector< T > &v ) {os << "{"; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << *it;} return os << "}";}
template < typename T >ostream &operator << ( ostream & os, const set< T > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin()) os << ", "; os << *it;} return os << "]";}
template < typename F, typename S >ostream &operator << ( ostream & os, const map< F, S > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << it -> first << " = " << it -> second ;} return os << "]";}
#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)
clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)
void faltu () { cerr << endl; }
template <typename T>void faltu( T a[], int n ) {for (int i = 0; i < n; ++i) cerr << a[i] << ' '; cerr << endl;}
template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }

// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
using namespace __gnu_pbds;

// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;
// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।
//*//**___________________________________________________**/
const int N = 56;
ll dp[N][N];

void calc()
{
    for (int i = 0; i < N; i++) {
        for (int j = 0; j <= i; j++) {
            if (j == 0 or i == j)
                dp[i][j] = 1ll;
            else
                dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]);
        }
    }
}
int main()
{
    //FASTIO
    /*
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif
//*/
    calc();
    int n, a, b;
    cin >> n >> a >> b;
    vector<ll> v(n);
    for (auto &x : v)cin >> x;

    sort(v.rbegin(), v.rend());

    double mx_avg = 0.0;
    for (int i = 0; i < a; i++) {
        mx_avg += v[i];
    }
    mx_avg /= a;

    int cntA = 0;
    int posA = 0;
    for (int i = 0; i < n; i++) {
        if (v[i] == v[a - 1]) {
            cntA++;
            if (i < a)posA++;
        }
    }

    ll cnt = 0ll;
    if (posA == a) {
        for (posA = a; posA <= b; posA++)
            cnt += dp[cntA][posA];
    }
    else {
        cnt += dp[cntA][posA];
    }

    cout.precision(20);
    cout << fixed << mx_avg << "\n";
    cout << cnt << "\n";
    return 0;
}