題目描述
把n個骰子扔在地上,所有骰子朝上一面的點數之和爲s。輸入n,打印出s的所有可能的值出現的概率。
你需要用一個浮點數數組返回答案,其中第 i 個元素代表這 n 個骰子所能擲出的點數集合中第 i 小的那個的概率。
示例 1:
輸入: 1
輸出: [0.16667,0.16667,0.16667,0.16667,0.16667,0.16667]
示例 2:
輸入: 2
輸出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111,0.08333,0.05556,0.02778]
思路實現
class Solution {
public double[] twoSum(int n) {
int [][]dp = new int[n+1][6*n+1];
//邊界條件
for(int s=1;s<=6;s++)dp[1][s]=1;
for(int i=2;i<=n;i++){
for(int s=i;s<=6*i;s++){
//求dp[i][s]
for(int d=1;d<=6;d++){
if(s-d<i-1)break;//爲0了
dp[i][s]+=dp[i-1][s-d];
}
}
}
double total = Math.pow((double)6,(double)n);
double[] ans = new double[5*n+1];
for(int i=n;i<=6*n;i++){
ans[i-n]=((double)dp[n][i])/total;
}
return ans;
}
}