gdcpc2015A題

題目名稱:Article

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=5236（在hdu的地址）

Problem Description

As the term is going to end, DRD begins to write his final article.

DRD uses the famous Macrohard's software, World, to write his article. Unfortunately this software is rather unstable, and it always crashes. DRD needs to write ncharacters in his article. He can press a key to input a character at time i+0.1, where i is an integer equal or greater than 0. But at every time i−0.1 for integer istrictly greater than 0, World might crash with probability p and DRD loses his work, so he maybe has to restart from his latest saved article. To prevent write it again and again, DRD can press Ctrl-S to save his document at time i. Due to the strange keyboard DRD uses, to press Ctrl-S he needs to press x characters. If DRD has input his total article, he has to press Ctrl-S to save the document. 

Since World crashes too often, now he is asking his friend ATM for the optimal strategy to input his article. A strategy is measured by its expectation keys DRD needs to press. 

Note that DRD can press a key at fast enough speed. 

 

Input

First line: an positive integer 0≤T≤20 indicating the number of cases.
Next T lines: each line has a positive integer n≤105, a positive real 0.1≤p≤0.9, and a positive integer x≤100.

 

Output

For each test case: output ''Case #k: ans'' (without quotes), where k is the number of the test cases, and ans is the expectation of keys of the optimal strategy.
Your answer is considered correct if and only if the absolute error or the relative error is smaller than 10−6. 

 

Sample Input

2 1 0.5 2 2 0.4 2

 

Sample Output

Case #1: 4.000000 Case #2: 6.444444

 

思路：dp ，用f[i]表示打出前i個字符所用時的數學期望

         轉移方程爲：f[i] = min (f[j] + g[i-j] + x)

         其中g[i]表示連續打印i個字符不出錯的期望

         所以可以推出g[i] =(g[i-1] + 1) * (1 - p) + (g[i-1] + 1 + g[i]) * p ，前半部分求到i不出錯的，後面求到i出錯的，出錯就得重新開始

         所以g[i] = (g[i-1] + 1) / (1 - p)

 不難發現，g的增長速率非常快，故轉移時只需考察前面很少的狀態

代碼如下：

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=100000+5;
double f[maxn];
int main()
{
    int t,n,x,cnt=0;
    double p;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%lf%d",&n,&p,&x);
        f[0]=0;
        for(int i=1;i<=n;i++)
        {
            double g=1/(1-p);
            f[i]=f[i-1]+g+x;
            for(int j=i-2;j>=0;j--)
            {
                g=(g+1)/(1-p);
                if(g>f[i])
                    break;
                f[i]=min(f[i],f[j]+g+x);
            }
        }
        printf("Case #%d: %.6f\n",++cnt,f[n]);
    }
    return 0;
}