"bzoj2318" Spoj4060 game with probability problem

Title Description

Alice and Bob are playing a game. There are n stones here, Alice and Bob take turns throwing coins, if the face up, then remove a stone from n stones, otherwise do nothing. The man who takes the last stone wins. Alice throws a shot at the probability of a p when she throws a coin, and, similarly, Bob has the probability of Q to throw out his congenial side.

Now Alice throws a coin, assuming they all want to win the game and ask you what the probability of Alice winning is.

Input

The first line is a positive integer t, which represents the number of data groups.

For each set of data, a row of three n,p,q.

Output

for each set of data output a single real number, which represents the probability of Alice winning, retains 6 decimal places.

Sample input

1
1 0.5 0.5

Sample output

0.666667

Tips

Data range:1<=t<=500.5<=p,q<=0.99999999for 100% of data 1<=n<=99999999 SolvingProbability DP The problem is really huge pit ... F[i] represents the probability that the first cast of a stone will win, and G[i] represents the probability of the winner of the I-block stone. Easy to launch f[i]= (p0*g[i-1]+ (1-p0) *q0*f[i-1])/(n (1-p0) * (1-q0)) g[i]= (q0*f[i-1]+ (1-q0) *p0*g[i-1])/(n (1-p0) * (1-q0)) here P Both 0 and q0 are target probabilities, and P and Q are the odds, so you need to decide if you want to face up. When F[i-1]<g[i-1], all want to face up, p0=p,q0=q, when F[i-1]>g[i-1], do not want to face up, p0=1-p,q0=1-q. But is n too big to be done? So using the probability of black technology: when n is getting bigger, f[n] gradually approaching a fixed value, and the topic only requires the retention of 6 decimal places. So the question F[1000+k] can be approximately equal to f[1000]. The time complexity is then reduced to O (1000T), which can be solved.

1#include <cstdio>2#include <cstring>3 Doublef[1001], g[1001];4 intMain ()5 {6     intT;7scanf"%d", &t);8      while(T-- )9     {Ten         intn, I; One         Doublep, q; Ascanf"%D%LF%LF", &n, &p, &q); -Memset (F,0,sizeof(f)); -Memset (G,0,sizeof(g)); the         if(N > +) -n = +; -f[0] =0; -g[0] =1; +          for(i =1; I <= N; i + + ) -         { +             if(F[i-1] > G[i-1]) Ap =1-P, q =1-Q; atF[i] = (p * g[i-1] + (1-P) * Q * f[i-1]) / (1- (1-P) * (1-q)); -G[i] = (Q * f[i-1] + (1-q) * p * g[i-1]) / (1- (1-P) * (1-q)); -             if(F[i-1] > G[i-1]) -p =1-P, q =1-Q; -         } -printf"%.6lf\n", F[n]); in     } -     return 0; to}

"bzoj2318" Spoj4060 game with probability problem