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docs/notes/60. n 个骰子的点数.md

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+# 60. n 个骰子的点数
+
+[Lintcode](https://www.lintcode.com/en/problem/dices-sum/)
+
+## 题目描述
+
+把 n 个骰子扔在地上，求点数和为 s 的概率。
+
+<img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/195f8693-5ec4-4987-8560-f25e365879dd.png" width="300px">
+
+## 解题思路
+
+### 动态规划
+
+使用一个二维数组 dp 存储点数出现的次数，其中 dp[i][j] 表示前 i 个骰子产生点数 j 的次数。
+
+空间复杂度：O(N<sup>2</sup>)
+
+```java
+public List<Map.Entry<Integer, Double>> dicesSum(int n) {
+    final int face = 6;
+    final int pointNum = face * n;
+    long[][] dp = new long[n + 1][pointNum + 1];
+
+    for (int i = 1; i <= face; i++)
+        dp[1][i] = 1;
+
+    for (int i = 2; i <= n; i++)
+        for (int j = i; j <= pointNum; j++)     /* 使用 i 个骰子最小点数为 i */
+            for (int k = 1; k <= face && k <= j; k++)
+                dp[i][j] += dp[i - 1][j - k];
+
+    final double totalNum = Math.pow(6, n);
+    List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
+    for (int i = n; i <= pointNum; i++)
+        ret.add(new AbstractMap.SimpleEntry<>(i, dp[n][i] / totalNum));
+
+    return ret;
+}
+```
+
+### 动态规划 + 旋转数组
+
+空间复杂度：O(N)
+
+```java
+public List<Map.Entry<Integer, Double>> dicesSum(int n) {
+    final int face = 6;
+    final int pointNum = face * n;
+    long[][] dp = new long[2][pointNum + 1];
+
+    for (int i = 1; i <= face; i++)
+        dp[0][i] = 1;
+
+    int flag = 1;                                     /* 旋转标记 */
+    for (int i = 2; i <= n; i++, flag = 1 - flag) {
+        for (int j = 0; j <= pointNum; j++)
+            dp[flag][j] = 0;                          /* 旋转数组清零 */
+
+        for (int j = i; j <= pointNum; j++)
+            for (int k = 1; k <= face && k <= j; k++)
+                dp[flag][j] += dp[1 - flag][j - k];
+    }
+
+    final double totalNum = Math.pow(6, n);
+    List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
+    for (int i = n; i <= pointNum; i++)
+        ret.add(new AbstractMap.SimpleEntry<>(i, dp[1 - flag][i] / totalNum));
+
+    return ret;
+}
+```