Lightoj 1248 機率dp

Problem:

給你 n 個骰子,求其抛擲它,直到所有面都出現過的抛擲次數期望值.

Analyse:

- 定義dp[i][j]為抛擲 j 個骰子,已經出現過i個不同的數字的機率,有轉移方程:

dp[i][j]=ij ∗(1+dp[i][j]) + j−ij∗(1+dp[i+1][j])

化簡發現與 j <script type="math/tex" id="MathJax-Element-24">j</script>無關,就是一個一維dp…

/**********************jibancanyang**************************
 *Author*        :jibancanyang
 *Created Time*  : 三  5/11 17:09:58 2016
 *File Name*     : jy.cpp

**Code**:
***********************[email protected]**********************/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
vector<int> vi;
#define pr(x) cout << #x << ": " << x << "  " 
#define pl(x) cout << #x << ": " << x << endl;
#define pri(a) printf("%d\n",(a));
#define xx first
#define yy second
#define sa(n) scanf("%d", &(n))
#define sal(n) scanf("%lld", &(n))
#define sai(n) scanf("%I64d", &(n))
#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) 
const int mod = int() + , INF = ;
const int maxn =  + ;
double dp[maxn];
int n;


int main(void)
{
#ifdef LOCAL
    //freopen("/Users/zhaoyang/in.txt", "r", stdin);
    //freopen("/Users/zhaoyang/out.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false),cin.tie(),cout.tie();
    int T; sa(T);
    for (int cas = ; cas <= T; cas++) {
        sa(n);
        dp[n] = ;
        for (int i = n - ; i >= ; i--) dp[i] = double(n) / (n - i) + dp[i + ];
        printf("Case %d: %.15f\n", cas, dp[]);
    }

    return ;
}