#20 Dices Sum

题目描述：

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

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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;        }    }    */};