while play load space n-1 acc amount blog image

Here’s a jolly and simple game: line up a row of N identical coins, all with the heads facing down onto the table and the tails upwards, and for exactly K times take one of the coins, toss it into the air, and replace it as it lands either heads-up or heads-down. You may keep all of the coins that are face-up by the end. Being, as we established last year, a ruthless capitalist, you have resolved to play optimally to win as many coins as you can. Across all possible combinations of strategies and results, what is the maximum expected (mean average) amount you can win by playing optimally? Input

One line containing two space-separated integers: 　　N(1<=N<=400), the number of coins at your mercy; 　　K(1<=K<=400), the number of flips you must perform. Output Output the expected number of heads you could have at the end, as a real number. The output must be accurate to an absolute or relative error of at most 10^-6.

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題意：給出N個硬幣，開始均反面朝上。每次挑出其中一個拋，連續K次，求正面朝上的最大數學期望。
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由於是求最大數學期望，所以每次拋硬幣即要優先選擇反面硬幣

所以只有兩種挑選硬幣的情況：

　　1.正面數量為 0 ~ n-1 ，選擇反面硬幣拋，拋出結果正面數量比原本 +1 或 不變

　　2.正面數量為 n，只能夠選擇正面硬幣拋，拋出結果正面數量比原本 -1 或 不變
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設 dp[i][j] 表示： 第 i 次拋硬幣後， j 個硬幣正面朝上的概率

　　1.當 j < n 時，dp[i][j]的概率一分為二，各給dp[i+1][j]和dp[i+1][j+1]，即

for(int j=0;j<n;j++){
    dp[i+1][j]+=dp[i][j]/2;
    dp[i+1][j+1]+=dp[i][j]/2;
}

　　2.當 j == n 時，dp[i][j]的概率一分為二，各給dp[i+1][j]和dp[i+1][j-1]，即

dp[i+1][n]+=dp[i][n]/2;
dp[i+1][n-1]+=dp[i][n]/2;

如此即可求出n個硬幣拋k次的各個正面朝上的概率，最後求數學期望即可

附：n=2，k=4時的dp轉移表格：

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <queue>
 7 #include <stack>
 8 #include <functional>
 9 #define INF 0x3f3f3f3f
10 using namespace std;
11 typedef long long ll;
12 double dp[410][410], ans;
13 int main(){
14     int n, k;
15     while (~scanf("%d %d", &n, &k)){
16         for (int i = 0; i <= 400; i++)
17             for (int j = 0; j <= 400; j++)
18                 dp[i][j] = 0;
19         dp[0][0] = 1;
20         for (int i = 0; i < k; i++){
21             for (int j = 0; j < n; j++){
22                 dp[i + 1][j] += dp[i][j] * 0.5;
23                 dp[i + 1][j + 1] += dp[i][j] * 0.5;
24             }
25             dp[i + 1][n] += dp[i][n] * 0.5;
26             dp[i + 1][n - 1] += dp[i][n] * 0.5;
27         }
28         ans = 0;
29         for (int i = 1; i <= n; i++)
30             ans += i*dp[k][i];
31         printf("%.8lf\n", ans);
32     }
33     return 0;
34 }

AC代碼

拋硬幣 Flipping Coins(Gym - 101606F)