One Person Game Time Limit: 1 Second      Memory Limit: 32768 KB      Special Judge

There is a very simple and interesting one-person game. You have 3 dice, namelyDie1, Die2 and Die3. Die1 hasK₁ faces. Die2 has K₂ faces.Die3 has K₃ faces. All the dice are fair dice, so the probability of rolling each value, 1 toK₁

, K₂, K₃ is exactly 1 /K₁, 1 / K₂ and 1 / K₃. You have a counter, and the game is played as follow:

1. Set the counter to 0 at first.
2. Roll the 3 dice simultaneously. If the up-facing number of Die1 isa, the up-facing number of Die2 is b and the up-facing number ofDie3 is c, set the counter to 0. Otherwise, add the counter by the total value of the 3 up-facing numbers.
3. If the counter's number is still not greater than n, go to step 2. Otherwise the game is ended.

Calculate the expectation of the number of times that you cast dice before the end of the game.

Input

There are multiple test cases. The first line of input is an integer T (0 <T <= 300) indicating the number of test cases. Then T

test cases follow. Each test case is a line contains 7 non-negative integersn, K₁, K₂, K₃,a, b, c (0 <= n <= 500, 1 < K₁,K₂, K₃ <= 6, 1 <= a <= K₁, 1 <=b <= K₂, 1 <= c <= K₃).

Output

For each test case, output the answer in a single line. A relative error of 1e-8 will be accepted.

Sample Input

2
0 2 2 2 1 1 1
0 6 6 6 1 1 1

Sample Output

1.142857142857143
1.004651162790698

題意：3個骰子，分別為k1,k2,k3面，每次三個骰子擲到a,b,c時，分數置為0,否則將分數加到計數器上，當計數器的分數大於n時，遊戲結束，問遊戲 結束的步數的數學期望

DP[i] = SUM(DP[i+k]*Pk)+DP[0]*p0+1

因為要求的是DP[0] 因此可以看成 一個常數

DP[i] = A[i]*DP[0] + B[i]

將DP[i+k]帶入式子1,化簡得

DP[i] = DP[0]*（sum(A[i+k])*Pk+p0）+sum(B[i+k]*Pk)+1

得出

A[i] = sum(A[i+k])*Pk+p0

B[i] = sum(B[i+k])*Pk+1

DP[0] = B[0]/(1-A[0])

算概率的時候要注意a,b,c的情況不能算入P（a+b+c）中

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 500+10;
int n,k1,k2,k3,a,b,c;
double A[maxn],B[maxn];
double P[20];
int main(){

    int ncase;
    cin >> ncase;
    while(ncase--){
        scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
        double p0 = 1.0/(k1*k2*k3);
        for(int i = 3; i <= k1+k2+k3; i++)
            P[i] = 0.0;
        for(int i = 1; i <= k1; i++)
            for(int j = 1; j <= k2; j++)
                for(int k = 1; k <= k3; k++)
                    if(!(i==a&&j==b&&k==c))
                        P[i+j+k] += p0;

        memset(A,0,sizeof  A);
        memset(B,0,sizeof B);
        for(int i = n; i >= 0; i--){
            A[i] = p0;
            B[i] =  1;
            for(int k = 3; k <= k1+k2+k3; k++){
                if(i+k<=n){
                    A[i] += A[i+k]*P[k];
                    B[i] += B[i+k]*P[k];
                }
            }
        }
        printf("%.8lf\n",B[0]/(1-A[0]));
    }
    return 0;
}