You are in a cave, a deep cave! The cave can be represented by an 1 x N grid. Some cells in the cave might contain gold!

Initially, you are in position 1. In each move, you throw a perfect 6 sided dice. If you get X in the dice after throwing, you add X to your position and collect all the gold from the new position. If your new position is outside of the cave, you keep throwing the dice again until you get a suitable result. When you reach the N^th position you stop your journey.

Given the information about the cave, you have to find the expected amount of gold you can collect using the procedure described above.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a blank line and an integer N (1≤ N ≤ 100) denoting the dimension of the cave. The next line contains N space separated integers. The i^th integer denotes the amount of gold you will get if you come to the i^th cell. You may safely assume that all the given integers will be non-negative and not greater than 1000.

Output

For each case, print the case number and the expected number of gold you will collect. Errors less than 10^-6 will be ignored.

Sample

Sample Input	Sample Output
Click to copy 3 1 101 2 10 3 3 3 6 9	Click to copy Case 1: 101 Case 2: 13.000000 Case 3: 15.0000000000

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LOJ 1030 - Discovering Gold

Discussion:

Given a 1XN grid, each cell contains some gold. In the beginning, you stand at position 1. You have a perfect 6 sided dice, if after throwing you get a result x, then you will go x cell ahead and collect gold in it.
You have to find the expected amount of gold you can collect.

Solution Idea

The probability of getting 1, 2, 3, 4, 5 or 6 after throwing dice is 1/6
So, if you are now standing at position i, then the probability of going to i+j cell (where i<6) is 1/6. (1/6 if i+6 <= N)
So, if we stand at position i, then expected amount of gold we can collect is: (expected_value => Ex_v)
gold[i]+ ( 1/6*Ex_v(i+1) + 1/6*Ex_v(i+2) + 1/6*Ex_v(i+3) + 1/6*Ex_v(i+4) + 1/6*Ex_v(i+4) + 1/6*Ex_v(i+6) )
It is true if i+6 <= N.
If i+6 > N then we have to replace 6 with N-i, as we don't have 6 cases then.
We have a recurrence relation, now based on this we can write a dp solution.

Solution Code(C++)

#include <bits/stdc++.h>
using namespace std;

int n; 
int arr[200];
long double dp[200];

long double solve(int pos){
    if(pos == n) return arr[n];
    if(dp[pos] >= 0) return dp[pos];

    long double v = 0;
    for(int i=1; i<=6; i++){
        if(pos+i <= n){
            int dv = min(6, n - pos);
            v += (1.0/dv) * solve(pos+i);
        }
    }
    dp[pos] = arr[pos] + v;
    return dp[pos];
}
int main() {
    int t, cs = 1; cin>>t;
    while(cs <= t){
        cin>>n;
        for(int i=1; i<=n; i++){
            cin>>arr[i];
        }
        for(int i=0; i<=110; i++) dp[i] = -1;
        cout << fixed << setprecision(9);
        cout<<"Case "<<cs<<": "<<solve(1)<<endl;
        cs++;
    }
}

Tutorial source