剑指 Offer 60. n个骰子的点数
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
解析:
dp[i][j]表示i个骰子,和为j的情况数
则dp[i][j] = sum(dp[i - 1][j - k]) 其中k 属于[1, j - 1]
class Solution { public: vector<double> dicesProbability(int n) { vector<double> res(n * 6 - n + 1); //> dp[i][j]: i - 1个骰子,点数和为 j 的情况的数量 vector<vector<int>> dp(n + 1, vector<int>(n * 6 + 1, 0)); //> 初始化 dp[1][i] = 1; //> 1个骰子,点数和为[1,6]的情况只有一种 for (int i = 1; i <= 6; ++i) { dp[1][i] = 1; } //> 遍历顺序:第2个骰子 ---> 第n个骰子 for (int i = 2; i <= n; ++i) { //> j 为i个骰子的点数和情况 for (int j = i; j <= 6 * n; ++j) { for (int k = 1; k <= 6; ++k) { if(j-k > 0) dp[i][j] += dp[i-1][j-k]; else break; } } } double denominator = pow(6.0, n); for (int i = n,index = 0; i <= 6*n; ++i) { res[index++] = dp[n][i] / denominator; } return res; } };