Difficulties algorithm | 20. dice sum

Title Description

N throwing dice, and the sum of the numbers above is S.
Given n, S list all possible values and their corresponding probabilities.

Sample 1

输入：n = 1
输出：[[1, 0.17], [2, 0.17], [3, 0.17], [4, 0.17], [5, 0.17], [6, 0.17]]
解释：掷一次骰子，向上的数字和可能为1,2,3,4,5,6，出现的概率均为 0.17。

Sample 2

输入：n = 2
输出：[[2,0.03],[3,0.06],[4,0.08],[5,0.11],[6,0.14],[7,0.17],[8,0.14],[9,0.11],[10,0.08],[11,0.06],[12,0.03]]
解释：掷两次骰子，向上的数字和可能在[2,12]，出现的概率是不同的。

java solution to a problem

public class Solution {
    /**
     * @param n an integer
     * @return a list of Map.Entry<sum, probability>
     */
    public List<Map.Entry<Integer, Double>> dicesSum(int n) {
        // Write your code here
        // Ps. new AbstractMap.SimpleEntry<Integer, Double>(sum, pro)
        // to create the pair.
        List<Map.Entry<Integer, Double>> results = 
                new ArrayList<Map.Entry<Integer, Double>>();
        
        double[][] f = new double[n + 1][6 * n + 1];
        for (int i = 1; i <= 6; ++i)
            f[1][i] = 1.0 / 6;

        for (int i = 2; i <= n; ++i)
            for (int j = i; j <= 6 * n; ++j) {
                for (int k = 1; k <= 6; ++k)
                    if (j > k)
                        f[i][j] += f[i - 1][j - k];

                f[i][j] /= 6.0;
            }

        for (int i = n; i <= 6 * n; ++i) 
            results.add(new AbstractMap.SimpleEntry<Integer, Double>(i, f[n][i]));

        return results;
    }
}

C ++ solution to a problem

class Solution {
public:
    /**
     * @param n an integer
     * @return a list of pair<sum, probability>
     */
    vector<pair<int, double>> dicesSum(int n) {
        // Write your code here
        vector<pair<int, double>> results;
        vector<vector<double>> f(n + 1, vector<double>(6 * n + 1));

        for (int i = 1; i <= 6; ++i) f[1][i] = 1.0 / 6;
        
        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];
                f[i][j] /= 6.0;
            }

        for (int i = n; i <= 6 * n; ++i)
            results.push_back(make_pair(i, f[n][i]));

        return results;
    }
};

python problem solution

class Solution:
    # @param {int} n an integer
    # @return {tuple[]} a list of tuple(sum, probability)
    def dicesSum(self, n):
        # Write your code here
        results = []
        f = [[0 for j in xrange(6 * n + 1)] for i in xrange(n + 1)]
        
        for i in xrange(1, 7):
            f[1][i] = 1.0 / 6.0
        for i in xrange(2, n + 1):
            for j in xrange(i, 6 * n + 1):
                for k in xrange(1, 7):
                    if j > k:
                        f[i][j] += f[i - 1][j - k]
                f[i][j] /= 6.0

        for i in xrange(n, 6 * n + 1):
            results.append((i, f[n][i]))

        return results