【剑指 Offer 60. n个骰子的点数】

【剑指 Offer 60. n个骰子的点数】

1. 题目描述

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]

2. 思路

动态规划：

n个骰子共有n~6*n种点数。设i个骰子点数为j的个数为dp[i][j]

则dp[i][j] = ∑_k dp[i-1][j-k]

3. 代码

class Solution {
public:
    vector<double> dicesProbability(int n) {
        vector<int> dp(6*n+1, 0);
        for(int i = 1; i <= 6; i++) {
            dp[i] = 1;
        }
        for(int i = 2; i <= n; i++) {
            for(int j = i * 6; j >= i; j--) {
                dp[j] = 0;
                for(int cur = 1; cur <= 6; cur++) {
                    if(j - cur < i - 1)
                        break;
                    dp[j] += dp[j-cur];
                }
            }
        }
        vector<double> res(5*n + 1);
        double base = pow(6, n);
        for(int i = n; i <= 6 * n; i++) {
            res[i - n] = (dp[i] / base);
        }
        return res;
    }
};

 

一个关于dp的教程：https://mp.weixin.qq.com/s/Ef73zZv6wiaXwiJRnCLpoQ