lintcode 20 Dices Sum
阿新 • • 發佈:2018-11-09
題目描述
Description
Throw n dices, the sum of the dices’ faces is S. Given n, find the all possible value of S along with its probability.
- 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]].
思路
看到這題,首先想到的就是動態規劃了,二維陣列記錄概率,然後用公式來求解:
n是骰子數量,a是點數,f(n,a)指的是n個骰子點數為a的概率
有了公式就很方便進行求解了
- 第一層迴圈,表示骰子數量
- 第二層迴圈,表示點數
- 第三層迴圈,用來計算概率
程式碼
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<>();
double[][] dp = new double[n + 1][n * 6 + 1];
for (int i = 1; i < 7; i++) {
dp[1][i] = 1.0 / 6;
}
for (int i = 2; i <= n; i++) {
for (int j = i; j <= 6 * i; j++) {
for (int m = 1; m <= 6; m++) {
if (j > m) {
dp[i][j] += dp[i - 1][j - m];
}
}
dp[i][j] /= 6.0;
}
}
for (int i = n; i <= 6 * n; ++i) {
result.add(new AbstractMap.SimpleEntry<>(i, dp[n][i]));
}
return result;
}