Lintcode20 Dices Sum solution 题解

【题目描述】

Throw n dices, the sum of the dices‘ faces is S. Given n, find the all possible value of S along with its probability.

Notice:You do not care about the accuracy of the result, we will help you to output results.

扔 n 个骰子，向上面的数字之和为 S。给定 Given n，请列出所有可能的 S 值及其相应的概率。

注意:你不用关注答案的准确性，我们会帮你输出答案

【题目链接】

http://www.lintcode.com/en/problem/dices-sum/

【题目解析】

这题用dfs做感觉更加直观，但是过不了time cost。换成dp的方法我是这么想的：

用dp[i][j]表示有i + 1个骰子的情况下，掷到的和为j的次数。那么intialize这个dp[0][j], j = 1...6的值都为1，然后从i = 1开始做循环。i个骰子和i + 1个骰子的差别就是1个骰子（废话），所以再用一个k = 1...6进行遍历，那么i + 1个骰子掷到j + k的次数就是原来dp[i][j + k]的次数加上dp[i - 1][j]。

这样我们就求得了n个骰子的情况下，每个S出现的次数dp[n - 1][j], j = n...6 * n。那么概率就是每个dp[n - 1][j]除以出现的总次数sum(dp[n - 1][j]).

这里要注意dp的值可能很大，所以要用到long long，否则在有些test case（e.g., n = 15）的情况下，会出现负数答案。

【答案链接】

http://www.jiuzhang.com/solution/dices-sum/

时间： 2024-10-16 02:18:14

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