Hdu 1204 candy war (expectation of Markov Chains)
Candy wars
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 2336 Accepted Submission (s): 797
Problem Description on the evening of the end of the birthday Party, there were some sweets left, And Gandon wanted to take all of them away. Speakless said, "Yes, but we can play around for 24 o'clock, didn't you get some candy? In this way, if anyone wins a game, they will take the sugar from the other party until all the sugar from the other party is obtained ." If no one can calculate and the other party cannot calculate, no one will win, but if both parties can calculate or cannot, even if the draw is made, there will be no gains or losses of candy.
Speakless is a person who prefers to think about problems in advance. Since he initiated the candy war, he naturally wants to win (or else he will be able to win -_-). Now he needs your help to give you the probability of winning each game and the probability of winning each game by Gardon. Please give him the probability of winning this war.
Each Input row has four numbers, the number of sweets on the Speakless hand N, and the number of sweets on the Gardon hand M (0 <= N, M <= 50) probability p that can be answered by a single Speakless, and the probability q That Gardon can answer (0 <= p, q <= 1 ).
One number per line in Output indicates the probability that Speakless can win (expressed in percentage, which is retained to the second digit after the decimal point ).
Sample Input
50 50 0.5 0.510 10 0.51 0.550 50 0.51 0.5
Sample Output
0.500.600.88
F [I] indicates the probability that Speakless will win when I have a candy, then f [I] = p (1-q) * f [I + 1] + q (1-p) * f [I-1] + (1-p * (1-q)-q * (1-p) * f [I]; this equation indicates that the expected result of the current State is equal to the sum of the expected results of all States that the current State can reach. Simplified, (f [I + 1]-f [I-1]) * p * (1-q) = f [I]-f [I-1] * (q * (1-p ));
K = (q * (1-p)/(p * (1-q ));
Set f [I + 1]-f [I] = (f [I]-f [I-1]) * k;
F [0] = 0; f [n + m] = 1; (mandatory and mandatory)
F [1]-f [0] = f [1]; f [2]-f [1] = (f [1]) * k; f [3]-f [2] = (f [2]-f [1]) * k = f [1] * k ^ 2 ;... f [n]-f [n-1] = (f [n-1]-f [n-1]) * k = f [1] * k ^ (n-1); summation of the series, f [n] = (1-k ^ n)/(1-k); f [n + m] = (1-k ^ (n + m)/(1-k) = 1;
Then f [n] = f [n]/f [n + m] = (1-k ^ n)/(1-k ^ (n + m ));
# Include
# Include
# Include
# Include
# Include
Using namespace std; # define ll long # define N 25 const int M = 256; const double eps = 1e-4; int main () {int n, m, n1, n2, i; double k, s, p, q; while (~ Scanf ("% d % lf", & n, & m, & p, & q) {if (m = 0) s = 1; else if (n = 0 | p = 0 | q = 1) s = 0; else {k = q * (1-p)/p/(1-q ); if (fabs (k-1.0)
