zoj3329--One Person Game(機率dp第六彈：形成環的dp，帶入係數，高斯消元)，zoj3329--onedp

zoj3329--One Person Game(機率dp第六彈：形成環的dp，帶入係數，高斯消元)，zoj3329--onedp

One Person GameTime Limit: 1 Second      Memory Limit: 32768 KB      Special Judge

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

題目大意：給出了k1,k2,k3三個篩子，當k1 == a k2 == b k3 == c時分數歸零，否則累加，問當總和到n以上需要的次數期望

狀態方程很好寫，dp[i]代表由i到n以上需要的次數，dp[i] = ∑（p[j]*dp[i+j]）+q*dp[0] + 1；p[j]代表擲出和為j的機率，q為歸零的機率，但是為問題出現了，在狀態方程中有dp[0]這是我們要求解的值。所以要帶入係數dpa[],dpb[]，dp[i] = dpa[i] + dpb[i]*dp[0] ;

最後求解出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;}