到达棋盘末端的预期移动次数|动态规划

给定一个长度为N的线性板，编号从1到N ，任务是找到到达板的^第N^个单元所需的预期移动次数，如果我们从编号为1 的单元开始，并且在每一步我们掷一个立方骰子决定下一步行动。此外，我们不能超出董事会的范围。请注意，预期的移动次数可以是小数。
例子：

Input: N = 8 
Output: 7 
p1 = (1 / 6) | 1-step -> 6 moves expected to reach the end 
p2 = (1 / 6) | 2-steps -> 6 moves expected to reach the end 
p3 = (1 / 6) | 3-steps -> 6 moves expected to reach the end 
p4 = (1 / 6) | 4-steps -> 6 moves expected to reach the end 
p5 = (1 / 6) | 5-steps -> 6 moves expected to reach the end 
p6 = (1 / 6) | 6-steps -> 6 moves expected to reach the end 
If we are 7 steps away, then we can end up at 1, 2, 3, 4, 5, 6 steps 
away with equal probability i.e. (1 / 6). 
Look at the above simulation to understand better. 
dp[N – 1] = dp[7] 
= 1 + (dp[1] + dp[2] + dp[3] + dp[4] + dp[5] + dp[6]) / 6 
= 1 + 6 = 7
Input: N = 10 
Output: 7.36111 
 

方法：这个问题可以用动态规划解决。为了解决这个问题，首先决定DP的状态。一种方法是使用当前单元格和第N^个单元格之间的距离来定义 DP 的状态。我们称这个距离为X 。因此DP [X]作为步骤的预期数量需要达到长度X + 1从^第一单元开始的基板的端部可以限定。
因此，递推关系变为：

dp[X] = 1 + (dp[X – 1] + dp[X – 2] + dp[X – 3] + dp[X – 4] + dp[X – 5] + dp[X – 6]) / 6 
 

现在，对于基本情况：

dp[0] = 0 
Let’s try to calculate dp[1]. 
dp[1] = 1 + 5 * (dp[1]) / 6 + dp[0] (Why? its because (5 / 6) is the probability it stays stuck at 1.) 
dp[1] / 6 = 1 (since dp[0] = 0) 
dp[1] = 6 
Similarly, dp[1] = dp[2] = dp[3] = dp[4] = dp[5] = 6 
 

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#define maxSize 50
using namespace std;
 
// To store the states of dp
double dp[maxSize];
 
// To determine whether a state
// has been solved before
int v[maxSize];
 
// Function to return the count
double expectedSteps(int x)
{
 
    // Base cases
    if (x == 0)
        return 0;
    if (x <= 5)
        return 6;
 
    // If a state has been solved before
    // it won't be evaluated again
    if (v[x])
        return dp[x];
 
    v[x] = 1;
 
    // Recurrence relation
    dp[x] = 1 + (expectedSteps(x - 1) +
                 expectedSteps(x - 2) +
                 expectedSteps(x - 3) +
                 expectedSteps(x - 4) +
                 expectedSteps(x - 5) +
                 expectedSteps(x - 6)) / 6;
    return dp[x];
}
 
// Driver code
int main()
{
    int n = 10;
 
    cout << expectedSteps(n - 1);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    static int maxSize = 50;
 
    // To store the states of dp
    static double dp[] = new double[maxSize];
     
    // To determine whether a state
    // has been solved before
    static int v[] = new int[maxSize];
     
    // Function to return the count
    static double expectedSteps(int x)
    {
     
        // Base cases
        if (x == 0)
            return 0;
             
        if (x <= 5)
            return 6;
     
        // If a state has been solved before
        // it won't be evaluated again
        if (v[x] == 1)
            return dp[x];
     
        v[x] = 1;
     
        // Recurrence relation
        dp[x] = 1 + (expectedSteps(x - 1) +
                     expectedSteps(x - 2) +
                     expectedSteps(x - 3) +
                     expectedSteps(x - 4) +
                     expectedSteps(x - 5) +
                     expectedSteps(x - 6)) / 6;
         
        return dp[x];
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 10;
     
        System.out.println(expectedSteps(n - 1));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
maxSize = 50
 
# To store the states of dp
dp = [0] * maxSize
 
# To determine whether a state
# has been solved before
v = [0] * maxSize
 
# Function to return the count
def expectedSteps(x):
 
    # Base cases
    if (x == 0):
        return 0
    if (x <= 5):
        return 6
 
    # If a state has been solved before
    # it won't be evaluated again
    if (v[x]):
        return dp[x]
 
    v[x] = 1
 
    # Recurrence relation
    dp[x] = 1 + (expectedSteps(x - 1) +
                 expectedSteps(x - 2) +
                 expectedSteps(x - 3) +
                 expectedSteps(x - 4) +
                 expectedSteps(x - 5) +
                 expectedSteps(x - 6)) / 6
    return dp[x]
 
# Driver code
n = 10
 
print(round(expectedSteps(n - 1), 5))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
    static int maxSize = 50;
 
    // To store the states of dp
    static double []dp = new double[maxSize];
     
    // To determine whether a state
    // has been solved before
    static int []v = new int[maxSize];
     
    // Function to return the count
    static double expectedSteps(int x)
    {
     
        // Base cases
        if (x == 0)
            return 0;
             
        if (x <= 5)
            return 6;
     
        // If a state has been solved before
        // it won't be evaluated again
        if (v[x] == 1)
            return dp[x];
     
        v[x] = 1;
     
        // Recurrence relation
        dp[x] = 1 + (expectedSteps(x - 1) +
                     expectedSteps(x - 2) +
                     expectedSteps(x - 3) +
                     expectedSteps(x - 4) +
                     expectedSteps(x - 5) +
                     expectedSteps(x - 6)) / 6;
         
        return dp[x];
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 10;
     
        Console.WriteLine(expectedSteps(n - 1));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Javascript implementation of the approach
var maxSize = 50;
 
// To store the states of dp
var dp = Array(maxSize);
 
// To determine whether a state
// has been solved before
var v = Array(maxSize);
 
// Function to return the count
function expectedSteps(x)
{
 
    // Base cases
    if (x == 0)
        return 0;
    if (x <= 5)
        return 6;
 
    // If a state has been solved before
    // it won't be evaluated again
    if (v[x])
        return dp[x];
 
    v[x] = 1;
 
    // Recurrence relation
    dp[x] = 1 + (expectedSteps(x - 1) +
                 expectedSteps(x - 2) +
                 expectedSteps(x - 3) +
                 expectedSteps(x - 4) +
                 expectedSteps(x - 5) +
                 expectedSteps(x - 6)) / 6;
    return dp[x];
}
 
// Driver code
var n = 10;
document.write( expectedSteps(n - 1).toFixed(5));
 
// This code is contributed by noob2000.

输出：

7.36111

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