Zoj 3329 DP expectation

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, 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 NumberDie1IsA, The up-facing NumberDie2IsBAnd the up-facing NumberDie3IsC, 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 greaterN, 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 integerT(0 <T<= 300) indicating the number of test cases. ThenTTest 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

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

Sample output

1.1428571428571431.00000000002790698
 

Double P [510]; // coefficient of DP [0] In DP [I]
Double CON [510]; // constant of DP [I]
// DP [I] = P [I] * DP [0] + con [I];
// DP [I] indicates the number of times to throw a sieve for I.
// List n equations
/*
DP [N] = DP [0] * x + sigma (DP [n + k] * sum [k]);
DP [n-1] = DP [0] * x + sigma (DP [n + k] * sum [k]);
.....
DP [0] = DP [0] * x + sigma (DP [0 + k] * sum [k]);
Where k is the sum of the possible number of sieve, sum [k] is the probability of K

View code

# Include <cstdio>
# Include <cstring>
# Include <algorithm>
 Using   Namespace STD;
 Double Sum [ 100 ];
Double P [ 510 ]; //  Coefficient of DP [0] In DP [I]  
  Double CON [ 510 ]; //  Constants of DP [I]  
  Int Main ()
{
 Int T, n, K1, K2, K3, I, J, K, A, B, C;
Scanf ( " % D  " , & T );
 While (T --)
{
Scanf ( "  % D  " , & N, & K1, & K2, & K3, & A, & B, & C );
 Double X = ( 1.0 /K1 * 1.0 /K2 * 1.0 /K3 );
Memset (p, 0 ,Sizeof (P ));
Memset (con, 0 , Sizeof (CON ));
Memset (sum, 0 , Sizeof (SUM ));
 For (I = 1 ; I <= k1; I ++)
{
 For (J = 1 ; J <= k2; j ++)
{
 For (K = 1 ; K <= K3; k ++)
{
 If (I = A & J = B & K = C) Continue ;
Sum [I + J + k] + = X;
}
}
}
 For (I = N; I> = 0 ; I --)
{
 For (J = I + 3 ; J <= k1 + k2 + K3 + I; j ++)
{
P [I] + = P [J] * sum [J-I];
CON [I] + = CON [J] * sum [J-I];
}
P [I] + = X;
CON [I] + = 1.0 ;
}
Printf ( "  %. 15lf \ n  " , CON [ 0 ]/( 1.0 -P [ 0 ]);
}
}