[LintCode]Dices Sum

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

骰子投掷n次，求所有情况的和出现的概率

二维数组dp[i][j]保存投掷i次得到和为j的概率，当前位置的概率为当前投掷1 ~ 6的情况下的前序概率和

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>> res = new LinkedList();        double[][] dp = new double[n + 1][6 * n + 1];        for (int i = 1; i <= 6; i++) {            dp[1][i] = 1.0 / 6.0;        }        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) {                        dp[i][j] += dp[i - 1][j - k];                    }                }                dp[i][j] /= 6.0;            }        }        for (int i = n; i <= 6 * n; i++) {            res.add(new AbstractMap.SimpleEntry<Integer, Double>(i, dp[n][i]));        }        return res;    }}

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