2019-11-02 12:07:41 +08:00
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# 60. n 个骰子的点数
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[Lintcode](https://www.lintcode.com/en/problem/dices-sum/)
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## 题目描述
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把 n 个骰子扔在地上，求点数和为 s 的概率。
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2019-11-02 17:33:10 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/195f8693-5ec4-4987-8560-f25e365879dd.png" width="300px"> </div><br>
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2019-11-02 12:07:41 +08:00
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## 解题思路
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### 动态规划
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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。
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空间复杂度：O(N<sup>2</sup>)
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```java
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public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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final int face = 6;
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final int pointNum = face * n;
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long[][] dp = new long[n + 1][pointNum + 1];
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for (int i = 1; i <= face; i++)
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dp[1][i] = 1;
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for (int i = 2; i <= n; i++)
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for (int j = i; j <= pointNum; j++) /* 使用 i 个骰子最小点数为 i */
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for (int k = 1; k <= face && k <= j; k++)
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dp[i][j] += dp[i - 1][j - k];
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final double totalNum = Math.pow(6, n);
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List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
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for (int i = n; i <= pointNum; i++)
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ret.add(new AbstractMap.SimpleEntry<>(i, dp[n][i] / totalNum));
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return ret;
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}
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```
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### 动态规划 + 旋转数组
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空间复杂度：O(N)
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```java
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public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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final int face = 6;
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final int pointNum = face * n;
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long[][] dp = new long[2][pointNum + 1];
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for (int i = 1; i <= face; i++)
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dp[0][i] = 1;
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int flag = 1; /* 旋转标记 */
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for (int i = 2; i <= n; i++, flag = 1 - flag) {
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for (int j = 0; j <= pointNum; j++)
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dp[flag][j] = 0; /* 旋转数组清零 */
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for (int j = i; j <= pointNum; j++)
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for (int k = 1; k <= face && k <= j; k++)
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dp[flag][j] += dp[1 - flag][j - k];
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}
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final double totalNum = Math.pow(6, n);
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List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
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for (int i = n; i <= pointNum; i++)
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ret.add(new AbstractMap.SimpleEntry<>(i, dp[1 - flag][i] / totalNum));
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return ret;
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}
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```
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2019-11-02 17:33:10 +08:00
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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