Dices Sum 20
Question
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.
Example
Given n = 1, return [ [1, 0.17], [2, 0.17], [3, 0.17], [4, 0.17], [5, 0.17], [6, 0.17]].
Solution
用dp解决。dp[i][j]表示i个骰子一共得到j点的概率。要得到dp[i][j]可以考虑若最后一个筛子的点数为k(1~6),则前i-1个筛子一共得到的点数为j-k(因为i-1个筛子至少得到i-1点,所以j-k >= i - 1 => k <= j - i + 1)。所以只要把最后一个筛子为k的各种情况加起来最后再除以6即可(每多一个筛子概率要多除以一个6)。
状态函数:
dp[i][j] = dp[i-1][j-k] (i <= j <= 6 * i, k >= 1 && k <= j - i + 1 && k <= 6)
代码如下:
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>> result = new ArrayList<Map.Entry<Integer, Double>>();
if(n < 1){
return result;
}
//初始化n=1的情况
double[][] matrix = new double[n + 1][6 * n + 1];
for(int i = 1; i <= 6; i++){
matrix[1][i] = 1.0/6;
}
for(int i = 2; i <= n; i++){
//i个筛子至少得到i点,至多得到6 * i点
for(int j = i; j <= 6 * i; j++){
//k表示最后一个筛子能取的点数
for(int k = 1; k <= 6; k++){
if(k <= j - i + 1){
matrix[i][j] += matrix[i - 1][j - k];
}
}
//相对i-1个筛子多了一个筛子,因此加和的每一项都要除以6
matrix[i][j] /= 6.0;
}
}
for(int i = n; i <= 6 * n; i++){
result.add(new AbstractMap.SimpleEntry<Integer, Double>(i, matrix[n][i]));
}
return result;
}
}