#20 Dices Sum

最新推荐文章于 2020-07-19 05:02:07 发布

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最新推荐文章于 2020-07-19 05:02:07 发布

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分类专栏： lintcode 文章标签： lintcode dynamic programming

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本文介绍了一种使用动态规划算法解决掷多个骰子时不同点数出现概率的问题。通过初始化状态转移方程并逐步迭代计算，最终得到每种点数出现的概率。

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题目描述：

This problem is in a contest: Career Fair Warm Up I . Submit your code and see your ranking!

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

Have you met this question in a real interview?

Yes

Example

Given n = 1, return[ [1, 0.17], [2, 0.17], [3, 0.17], [4, 0.17], [5, 0.17], [6, 0.17]].

题目思路：

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。

Mycode（AC = 12ms）:

class Solution {
public:
    /**
     * @param n an integer
     * @return a list of pair<sum, probability>
     */
    vector<pair<int, double>> dicesSum(int n) {
        vector<pair<int, double>> ans;
        vector<vector<long long>> dp(n, vector<long long>(n * 6 + 1, 0));
        
        // initialize the dp for n == 1
        for (int i = 1; i <= 6; i++) {
            dp[0][i] = 1;
        }
        
        // for each dp[i][j + k], it is equal to the
        // number from dp[i - 1][j] + old dp[i][j + k]
        for (int i = 1; i < n; i++) {
            for (int j = i; j < i * 6 + 1; j++) {
                for (int k = 1; k <= 6; k++) {
                   dp[i][j + k] += dp[i - 1][j];
                }
            }
        }
        
        // get the total count of number occurs
        long long total_count = 0;
        for (int i = n; i < n * 6 + 1; i++) {
            if (dp[n - 1][i] > 0) {
                total_count += dp[n - 1][i];
            }
        }
        
        // get the probability
        for (int i = 1; i < n * 6 + 1; i++) {
            if (dp[n - 1][i] > 0) {
                pair<int, double> p(i, (double)dp[n - 1][i]/(double)total_count);
                ans.push_back(p);
            }
        }

        return ans;
    }
    
    /*
    vector<pair<int, double>> dicesSum(int n) {
        // Write your code here
        vector<pair<int, double>> ans;
        vector<double> counter(n * 6 + 1, 0.0);

        dicesSumHelper(counter, n, 0);

        int total_count = 0;
        for (int i = 1; i < counter.size(); i++) {
            if (counter[i] > 0.0) {
                total_count += counter[i];
            }
        }

        for (int i = 1; i < counter.size(); i++) {
            if (counter[i] > 0.0) {
                pair<int, double> p(i, counter[i]/total_count);
                ans.push_back(p);
            }
        }

        return ans;
    }

    void dicesSumHelper(vector<double>& counter, int n, int sum) {
        if (n < 0) return;
        else if (n == 0) {
            counter[sum] += 1.0;
            return;
        } else {
            for (int i = 1; i <= 6; i++) {
                dicesSumHelper(counter, n - 1, sum + i);
            }
            return;
        }
    }
    */
};