light oj 1248 - Dice (III)（概率dp）

1248 - Dice (III)

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Time Limit: 1 second(s)

Memory Limit: 32 MB

Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability of occurring any face is equal.

For example, for a fair two sided coin, the result is 3. Because when you first throw the coin, you will definitely see a new face. If you throw the coin again, the chance of getting the opposite side is 0.5, and the chance of getting the same side is 0.5. So, the result is

1 + (1 + 0.5 * (1 + 0.5 * ...))

= 2 + 0.5 + 0.5² + 0.5³ + ...

= 2 + 1 = 3

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10⁵).

Output

For each case, print the case number and the expected number of times you have to throw the dice to see all its faces at least once. Errors less than 10^-6 will be ignored.

Sample Input	Output for Sample Input
5 1 2 3 6 100	Case 1: 1 Case 2: 3 Case 3: 5.5 Case 4: 14.7 Case 5: 518.7377517640

 

一个均匀的骰子有n个面 投色子 要求最后要把骰子的每一面都看到了  求扔骰子次数的期望

扔骰子只会出现两种情况 一种是扔出了已经出现的面 一种就是扔出了新的面

dp[i]表示扔出i面需要次数的期望 则可以得到dp[i]=i/n*dp[i]+(n-i)/n*dp[i+1]+1

得到dp[i]=dp[i+1]+n/(n-i)   其中dp[n]=0  答案为dp[0]

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"< '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}
double dp[100010];

int main()
{
//    fread;
    int tc;
    scanf("%d",&tc);
    int cs=1;
    while(tc--)
    {
        int n;
        scanf("%d",&n);
        dp[n]=0.0;
        for(int i=n-1;i>=0;i--)
        {
            dp[i]=dp[i+1]+(n*1.0)/((double)(n-i));
        }
        printf("Case %d: %.7lf\n",cs++,dp[0]);
    }
    return 0;
}