There is a very simple and interesting one-person game. You have 3 dice, namely Die1, Die2 and Die3. Die1 has K₁ faces. Die2 has K₂ faces. Die3 has K₃ faces. All the dice are fair dice, so the probability of rolling each value, 1 to K₁, 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 is a, the up-facing number of Die2 is b and the up-facing number of Die3 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 integers n, 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

20 2 2 2 1 1 10 6 6 6 1 1 1

Sample Output

1.1428571428571431.004651162790698

题目大意：给出了k1,k2,k3三个筛子，当k1 == a k2 == b k3 == c时分数归零，否则累加，问当总和到n以上需要的次数期望

状态方程很好写，dp[i]代表由i到n以上需要的次数，dp[i] = ∑（p[j]*dp[i+j]）+q*dp[0] + 1；p[j]代表掷出和为j的概率，q为归零的概率，但是为问题出现了，在状态方程中有dp[0]这是我们要求解的值。所以要带入系数dpa[],dpb[]，dp[i] = dpa[i] + dpb[i]*dp[0] ;

最后求解出dp[0] = dpa[0] + dpb[0]*dp[0]，可以解除dp[0];

dp[i] = dpa[i] + dpb[i]*dp[0] = ∑（p[j]*dp[i+j]）+ q*dp[0]+1;

得到dpa[i] = ∑( p[j]*dpa[i+j] ) + 1 ;   dpb[i] = ∑( p[j]*dpb[i+j] ) + q ;

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;double p[20] , q , cnt ;double dpa[600] , dpb[600] ;int main(){    int t , n , m , i , j , k , l , k1 , k2 , k3 , a , b , c ;    scanf("%d", &t);    while(t--)    {        scanf("%d %d %d %d %d %d %d", &n, &k1, &k2, &k3, &a, &b, &c);        memset(p,0,sizeof(p));        memset(dpa,0,sizeof(dpa));        memset(dpb,0,sizeof(dpb));        p[a+b+c] = -1 ;        cnt = 0 ;        m = k1 + k2 + k3 ;        for(i = 1 ; i <= k1 ; i++)            for(j = 1 ; j <= k2 ; j++)                for(k = 1 ; k <= k3 ; k++)                {                    p[i+j+k] += 1.0 ;                    cnt += 1.0 ;                }        for(i = 3; i <= m ; i++)            p[i] /= cnt ;        q = 1.0 / cnt ;        for(i = n ; i >= 0 ; i--)        {            dpa[i] = 1.0 ; dpb[i] = q ;            for(j = 3 ; j <= k1+k2+k3 ; j++)            {                dpa[i] += p[j]*dpa[i+j] ;                dpb[i] += p[j]*dpb[i+j] ;            }        }        printf("%.10lf\n", dpa[0]/(1-dpb[0]));    }    return 0;}

zoj3329--One Person Game(概率dp第六弹：形成环的dp，带入系数，高斯消元)