#P17393. [ICPC 2018 Shenyang R] Sequences Generator
[ICPC 2018 Shenyang R] Sequences Generator
Problem Description
The RBM Second Generation of Dual Core Microprocessor Chip, also known as RBM2gDCMC, can generate a digital sequence of length . Each digit in a sequence provided by RBM2gDCMC is regarded as an integer between and in this problem.
Now I will show you the passcode for the email belonging to Gini Romety, which is a sequence of length with integers between and . You are asked to calculate the probabilities of the coincidence with Gini Romety's passcode for all consecutive subsequence of length in a sequence generated by RBM2gDCMC.
Input Format
The input contains several test cases, and the first line contains a positive integer indicating the number of test cases which is up to .
For each test case, the first line contains two integers and , satisfying , which are described as above.
The following lines describe the generating logic for all digits in a sequence built by RBM2gDCMC. The -th line of them contains two integers and , satisfying and , and following integers, denoted by , where and . These data indicate that for the -th digit the probability of being an integer in is zero, and the probability of being an integer in is .
The next line contains integers, denoted by , describing the passcode for Gini Romety's email, where .
We guarantee that the sum of in all test cases is no larger than .
Output Format
For each test case, output a line containing "Case #x:" (without quotes) at first, where is the test case number starting from .
After that, output lines such that the -th of them contains a real number indicating the probability of the coincidence for the passcode of Gini Romety's email and the subsequence of a sequence produced by RBM2gDCMC from the -th digit to the -th one with an absolute error of at most . Precisely speaking, assume that your answer is and the jury's answer is , your answer will be considered correct if , where means the absolute value of .
1
5 3
1 3 100000000 200000000 700000000
1 3 600000000 150000000 250000000
1 3 333333333 333333334 333333333
3 4 450000000 550000000
1 3 999999998 1 1
1 2 3
Case #1:
0.004999999995000
0.090000000180000
0.000000000000000
Hint
In the sample case, the probability matrix is
$$\begin{bmatrix} 0.100000000 & 0.200000000 & 0.700000000 & 0.000000000 & 0.000000000 \\ 0.600000000 & 0.150000000 & 0.250000000 & 0.000000000 & 0.000000000 \\ 0.333333333 & 0.333333334 & 0.333333333 & 0.000000000 & 0.000000000 \\ 0.000000000 & 0.000000000 & 0.450000000 & 0.550000000 & 0.000000000 \\ 0.999999998 & 0.000000001 & 0.000000001 & 0.000000000 & 0.000000000 \end{bmatrix}$$and thus the answers in the output are
- $p_{1, 1} p_{2, 2} p_{3, 3} = 0.100000000 \times 0.150000000 \times 0.333333333 = 0.004999999995000$,
- $p_{2, 1} p_{3, 2} p_{4, 3} = 0.600000000 \times 0.333333334 \times 0.450000000 = 0.090000000180000$,
- $p_{3, 1} p_{4, 2} p_{5, 3} = 0.333333333 \times 0.000000000 \times 0.000000001 = 0.000000000000000$
respectively.