2318:spoj4060 game with probability problem time Limit: 1 Sec Memory Limit: MB
Submit: 259 Solved: 108
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Aliceand theBobplaying a game. Have aNhere's a stone.Aliceand theBobToss the coin in turn, if the face is facing up, then fromNtake a stone out of a stone, or do nothing. The man who takes the last stone wins. Alicewhen throwing coins, there isPthe probability of throwing out the side he wanted to cast, likewise,Bobhave aQthe probability of throwing 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 numbers 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 Input1
1 0.5 0.5
Sample Output0.666667
HINT
Data range:
1<=t<=50
0.5<=p,q<=0.99999999
For 100% of data 1<=n<=99999999
Game theory + probability DP
This problem I do not know very well, the direct handling of the problems (that I did not read):
http://blog.csdn.net/Vmurder/article/details/46467899
If someone understands, welcome to share with me, thank you.
There is a good idea: N is very large when growth for the answer is very small, so the N and 100 to the minimum, you can reduce the number of operations.
#include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <cmath > #include <algorithm> #define F (I,j,n) for (int. i=j;i<=n;i++) #define D (i,j,n) for (int i=j;i>=n;i--) # Define ll long longusing namespace Std;int t,n;double p,q,f[105],g[105];int main () {scanf ("%d", &t); f[0]=0;g[0]=1; while (t--) {scanf ("%d%lf%lf", &n,&p,&q); N=min (n,100); F (i,1,n) {if (f[i-1]>g[i-1]) p=1-p,q=1-q;f[i]= (p*g[i-1]+ (1-p) *q*f[i-1])/(n (1-p) * (1-Q)) g[i]= (q*f[i-1]+ (1-q) *p *G[I-1])/(1-p) * (1-Q); if (f[i-1]>g[i-1]) p=1-p,q=1-q;} printf ("%.6lf\n", F[n]);} return 0;}
bzoj2318 Spoj4060 game with probability problem