Question

面试题 60. n 个骰子的点数

题目描述

把n个骰子扔在地上，所有骰子朝上一面的点数之和为s。输入n，打印出s的所有可能的值出现的概率。

 

你需要用一个浮点数数组返回答案，其中第 i 个元素代表这 n 个骰子所能掷出的点数集合中第 i 小的那个的概率。

 

示例 1:

输入: 1
输出: [0.16667,0.16667,0.16667,0.16667,0.16667,0.16667]

示例 2:

输入: 2
输出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111,0.08333,0.05556,0.02778]

 

限制：

1 <= n <= 11

解法

方法一：动态规划

我们定义 $f[i][j]$ 表示投掷 $i$ 个骰子，点数和为 $j$ 的方案数。那么我们可以写出状态转移方程：

$$ f[i][j] = \sum_{k=1}^6 f[i-1][j-k] $$

其中 $k$ 表示当前骰子的点数，$f[i-1][j-k]$ 表示投掷 $i-1$ 个骰子，点数和为 $j-k$ 的方案数。

最终我们需要求的是 $f[n][n \sim 6n]$ 的和，即投掷 $n$ 个骰子，点数和为 $n \sim 6n$ 的方案数之和。

注意到 $f[i][j]$ 的值只与 $f[i-1][j-k]$ 有关，因此我们可以使用滚动数组的方法将空间复杂度降低到 $O(6n)$。

时间复杂度 $O(n^2)$，空间复杂度 $O(6n)$。其中 $n$ 为骰子个数。

class Solution:
  def dicesProbability(self, n: int) -> List[float]:
    f = [[0] * (6 * n + 1) for _ in range(n + 1)]
    for j in range(1, 7):
      f[1][j] = 1
    for i in range(2, n + 1):
      for j in range(i, 6 * i + 1):
        for k in range(1, 7):
          if j - k >= 0:
            f[i][j] += f[i - 1][j - k]
    m = pow(6, n)
    return [f[n][i] / m for i in range(n, 6 * n + 1)]

class Solution {
  public double[] dicesProbability(int n) {
    int[][] f = new int[n + 1][6 * n + 1];
    for (int j = 1; j <= 6; ++j) {
      f[1][j] = 1;
    }
    for (int i = 2; i <= n; ++i) {
      for (int j = i; j <= 6 * i; ++j) {
        for (int k = 1; k <= 6; ++k) {
          if (j >= k) {
            f[i][j] += f[i - 1][j - k];
          }
        }
      }
    }
    double m = Math.pow(6, n);
    double[] ans = new double[5 * n + 1];
    for (int i = 0; i < ans.length; ++i) {
      ans[i] = f[n][n + i] / m;
    }
    return ans;
  }
}

class Solution {
public:
  vector<double> dicesProbability(int n) {
    int f[n + 1][6 * n + 1];
    memset(f, 0, sizeof f);
    for (int j = 1; j <= 6; ++j) {
      f[1][j] = 1;
    }
    for (int i = 2; i <= n; ++i) {
      for (int j = i; j <= 6 * i; ++j) {
        for (int k = 1; k <= 6; ++k) {
          if (j >= k) {
            f[i][j] += f[i - 1][j - k];
          }
        }
      }
    }
    vector<double> ans(5 * n + 1);
    double m = pow(6, n);
    for (int i = 0; i < ans.size(); ++i) {
      ans[i] = f[n][n + i] / m;
    }
    return ans;
  }
};

func dicesProbability(n int) (ans []float64) {
  f := make([][]int, n+1)
  for i := range f {
    f[i] = make([]int, 6*n+1)
  }
  for j := 1; j <= 6; j++ {
    f[1][j] = 1
  }
  for i := 2; i <= n; i++ {
    for j := i; j <= 6*i; j++ {
      for k := 1; k <= 6; k++ {
        if j >= k {
          f[i][j] += f[i-1][j-k]
        }
      }
    }
  }
  m := math.Pow(6, float64(n))
  for j := n; j <= 6*n; j++ {
    ans = append(ans, float64(f[n][j])/m)
  }
  return
}

/**
 * @param {number} n
 * @return {number[]}
 */
var dicesProbability = function (n) {
  const f = Array.from({ length: n + 1 }, () => Array(6 * n + 1).fill(0));
  for (let j = 1; j <= 6; ++j) {
    f[1][j] = 1;
  }
  for (let i = 2; i <= n; ++i) {
    for (let j = i; j <= 6 * i; ++j) {
      for (let k = 1; k <= 6; ++k) {
        if (j >= k) {
          f[i][j] += f[i - 1][j - k];
        }
      }
    }
  }
  const ans = [];
  const m = Math.pow(6, n);
  for (let j = n; j <= 6 * n; ++j) {
    ans.push(f[n][j] / m);
  }
  return ans;
};

public class Solution {
  public double[] DicesProbability(int n) {
    var bp = new double[6];
    for (int i = 0; i < 6; i++) {
      bp[i] = 1 / 6.0;
    }
    double[] ans = new double[]{1};
    for (int i = 1; i <= n; i++) {
      var tmp = ans;
      ans = new double[tmp.Length + 5];
      for (int i1 = 0; i1 < tmp.Length; i1++) {
        for (int i2 = 0; i2 < bp.Length; i2++) {
          ans[i1+i2] += tmp[i1] * bp[i2];
        }
      }
    }
    return ans;
  }
}

方法二

class Solution:
  def dicesProbability(self, n: int) -> List[float]:
    f = [0] + [1] * 6
    for i in range(2, n + 1):
      g = [0] * (6 * i + 1)
      for j in range(i, 6 * i + 1):
        for k in range(1, 7):
          if 0 <= j - k < len(f):
            g[j] += f[j - k]
      f = g
    m = pow(6, n)
    return [f[j] / m for j in range(n, 6 * n + 1)]

func dicesProbability(n int) []float64 {
  dp := make([]float64, 7)
  for i := 1; i <= 6; i++ {
    dp[i] = 1.0 / 6.0
  }
  for i := 2; i <= n; i++ {
    n := len(dp)
    tmp := make([]float64, 6*i+1)
    for j := 0; j < n; j++ {
      for k := 1; k <= 6; k++ {
        tmp[j+k] += dp[j] / 6.0
      }
    }
    dp = tmp
  }
  return dp[n:]
}