2018CCPC吉林赛区 D

Time limit1000 ms Memory limit262144 kB Special judgeYes OSWindows

The Moon card shows a large, full moon in the night’s sky, positioned between two large towers. The Moon is a symbol of intuition, dreams, and the unconscious. The light of the moon is dim, compared to the sun, and only vaguely illuminates the path to higher consciousness which winds between the two towers.

Random Six is a FPS game made by VBI(Various Bug Institution). There is a gift named "Beta Pack". Mr. K wants to get a beta pack. Here is the rule.
Step 0. Let initial chance rate qq = 2%.
Step 1. Player plays a round of the game with winning rate pp.
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 qq. If he doesn’t get it, let qq = min(100%, qq + 2%) and he will go to Step 1.
Step 4. Let qq = min(100%, qq + 1.5%) and goto Step 1.
Mr. K has winning rate pp% , he wants to know what’s the expected number of rounds before he needs to play.

Input

The first line contains testcase number TT (TT ≤ 100). For each testcase the first line contains an integer pp (1 ≤ pp ≤ 100).OutputFor each testcase print Case ii : and then print the answer in one line, with absolute or relative error not exceeding 106106.

Sample Input

2
50
100

Sample Output

Case 1: 12.9933758002
Case 2: 8.5431270393

概率dp题，从100开始逆推，状态转移方程为dp[i]=p*((i/1000.0)+(1-i/1000.0)*(1+dp[min(i+20,1000)]))+(1-p)*(1+dp[min(1000,i+15)]);

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    for(int l=1;l<=t;++l)
    {
        double p;
        scanf("%lf",&p);
        double dp[1005];
        p/=100;
        dp[1000]=1/p;
        for(int i=999;i>=20;--i)
        {
            dp[i]=p*((i/1000.0)+(1-i/1000.0)*(1+dp[min(i+20,1000)]));
            dp[i]+=(1-p)*(1+dp[min(1000,i+15)]);
        }
        printf("Case %d: %.10f
",l,dp[20]);
    }
}