LightOJ1248---Dice (III)(概率dp)

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本文探讨了对于一个n面公平骰子，要至少看到每一面出现一次所需要的平均掷骰次数。通过动态规划的方法给出了计算这一期望值的具体算法，并提供了完整的C++实现代码。

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

100

Case 1: 1

Case 2: 3

Case 3: 5.5

Case 4: 14.7

Case 5: 518.7377517640

Problem Setter: Jane Alam Jan

水题概率dp

/*************************************************************************
    > File Name: g.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年04月30日 星期四 16时17分00秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

double dp[100100];
int n;

double dfs(int cur) {
    if (dp[cur] != -1.0) {
        return dp[cur];
    }
    double p = (n * 1.0 / (n - cur));
    dp[cur] = dfs(cur + 1) + p;
    return dp[cur];
}

int main() {
    int icase = 1, t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 0; i <= n; ++i) {
            dp[i] = -1.0;
        }
        dp[n] = 0;
        printf("Case %d: %.12f\n", icase++, dfs(0));
    }
}