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。
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12 | 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)]
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23 | 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;
}
}
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25 | 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;
}
};
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23 | 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
}
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25 | /**
* @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;
};
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19 | 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;
}
}
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方法二