LightOJ 1248 Dice (III) (水题，期望DP)

题意：给出一个n面的色子，问看到每个面的投掷次数期望是多少。

析：这个题很水啊，就是他解释样例解释的太。。。我鄙视他，，，，，

dp[i] 表示已经看到 i 面的期望是多少，然后两种选择一种是看到新的一面，另一种是看到旧的一面，然后就很出答案了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e16;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 10;

const int mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

double dp[maxn];

int main(){

  int T;  cin >> T;

  for(int kase = 1; kase <= T; ++kase){

    scanf("%d", &n);

    dp[1] = 1.0;

    for(int i = 2; i <= n; ++i)

      dp[i] = (dp[i-1] * (n - i + 1) + i - 1) / (n - i + 1) + 1.0;

    printf("Case %d: %.10f\n", kase, dp[n]);

  }

  return 0;

}

巴特西

LightOJ 1248 Dice (III) (水题，期望DP)

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