2018ccpc吉林 问题 D: THE MOON 概率dp

题意：

Step 0. Let initial chance rate q = 2%.
Step 1. Player plays a round of the game with winning rate p%.
Step 2. If the player wins, then will go to Step 3 else go to Step 4.
Step 3. Player gets a beta pack with probability q. If he doesn't get it, let q = min(100%, q + 2%) and he will go to Step 1.
Step 4. Let q = min(100%, q + 1.5%) and goto Step 1.
Mr. K has winning rate p% , he wants to know what's the expected number of rounds before he needs to play.

题解：dp[i][j]表示 q为i，p为j时的期望次数，因为有一个1.5的存在，所以我们的q都乘10进行计算

dp[q][p]=1+1.0*p/100*(1-1.0*q/1000)*dp[min(q+20,1000)][p]+(100.0-p)/100*dp[min(q+15,1000)][p]

初始dp[1000][i]=100/i

#include<bits/stdc++.h>
using namespace std;
double dp[1010][110];
double dfs(int q,int p)
{
	if(dp[q][p]!=-1)return dp[q][p];
	dp[q][p]=1+1.0*p/100*(1-1.0*q/1000)*dfs(min(q+20,1000),p)+(100.0-p)/100*dfs(min(q+15,1000),p);
	return dp[q][p];
}
int main()
{
	for(int i=1;i<=1000;i++)
	{
		for(int j=1;j<=100;j++)
			dp[i][j]=-1;
	}
	for(int i=1;i<=100;i++)
		dp[1000][i]=100.0/i;
	int T,p;
	int nn=1;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&p);
		printf("Case %d: %.10f\n",nn++,dfs(20,p));
	}
	return 0;
}

标签：min,int,MOON,2018ccpc,Step,100,dp,1000
来源： https://blog.csdn.net/mmk27_word/article/details/89789958