把n个骰子扔在地上,所有骰子朝上一面的点数之和为s。输入n,打印出s的所有可能的值出现的概率。
你需要用一个浮点数数组返回答案,其中第 i 个元素代表这 n 个骰子所能掷出的点数集合中第 i 小的那个的概率。
示例 1:
输入: 1
输出: [0.16667,0.16667,0.16667,0.16667,0.16667,0.16667]
示例 2:
输入: 2
输出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111,0.08333,0.05556,0.02778]
限制:
1 <= n <= 11
代码:
class Solution {
public double[] twoSum(int n) {
int dp[][] = new int[n+1][6*n+1];
double result[] = new double[5*n+1];
double x = Math.pow(6,n);
for(int i=1;i<=6;i++)
{
dp[1][i]=1;
}
for(int i=1;i<=n;i++)
{
for(int j=i;j<=6*n;j++)
{
for(int k=1;k<=6;k++)
{
if(j>=k)
{
dp[i][j]+=dp[i-1][j-k];
}
if(i==n)
{
result[j-i]=dp[i][j]/x;
}
}
}
}
return result;
}
}