BZOJ_P2318 SPOJ4060 game with probability Problem(概率动态规划)
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BZOJ传送门
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 224 Solved: 95
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Description
Alice和Bob在玩一个游戏。有n个石子在这里,Alice和Bob轮流投掷硬币,如果正面朝上,则从n个石子中取出一个石子,否则不做任何事。取到最后一颗石子的人胜利。Alice在投掷硬币时有p的概率投掷出他想投的一面,同样,Bob有q的概率投掷出他相投的一面。
现在Alice先手投掷硬币,假设他们都想赢得游戏,问你Alice胜利的概率为多少。
Input
第一行一个正整数t,表示数据组数。
对于每组数据,一行三个数n,p,q。
Output
对于每组数据输出一行一个实数,表示Alice胜利的概率,保留6位小数。
Sample Input
1
1 0.5 0.5
Sample Output
0.666667
HINT
数据范围:
1<=t<=50
0.5<=p,q<=0.99999999
对于100%的数据 1<=n<=99999999
Source
#include<cstdio>#include<cstring>#include<iostream>using namespace std;#define N 105int n,t;double p,q;double w1,w2,w3,w4,w5,w6,w7,w8;double f[N],g[N];int main(){ scanf("%d",&t); while(t--){ scanf("%d%lf%lf",&n,&p,&q); n=min(n,100);f[0]=0,g[0]=1.0; for(int i=1;i<=n;i++) { 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])/(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]); }}
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