题目链接 题意: 给一个质地均匀的n的骰子, 求投掷出所有点数至少一次的期望次数. 思路: 这就是一个经典的邮票收集问题(Coupon Collector Problem). 投掷出第一个未出现的点数的概率为n/n = 1, 因为第一次投掷必然是未出现的. 第二个未出现的点数第一次出现的概率为 (n - 1) / n,因为有一个已经投掷出现过. 第i个未出现的点数第一次出现的概率为 (n - i) / i, 这满足几何分布. 其期望E = 1/p 所以期望为n *(1 + 1 / 2 + 1 /…

G - Dice (III) Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Description Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that…

题目链接:https://vjudge.net/problem/LightOJ-1248 1248 - Dice (III)    PDF (English) Statistics Forum Time Limit: 1 second(s) Memory Limit: 32 MB Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see a…

题目链接:LightOJ - 1248 Description Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability…

1248 - Dice (III)   PDF (English) Statistics Forum Time Limit: 1 second(s) Memory Limit: 32 MB Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the di…

Description Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability of occurring any fa…

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; int t,n; double dp[100010]; int main() { scanf("%d",&t); int cas=1; while(t--) { scanf("%d",&n); dp[n]=0…

期望,$dp$. 设$dp[i]$表示当前已经出现过$i$个数字的期望次数.在这种状态下,如果再投一次,会出现两种可能,即出现了$i+1$个数字以及还是$i$个数字. 因此 $dp[i]=dp[i]*i/n+dp[i+1]*(n-i)/n+1$,即$dp[i]=dp[i+1]+n/(n-i)$,$dp[n]=0$,推出$dp[0]$即可. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1248 题意:有一个 n 面的骰子,问至少看到所有的面一次的所需 掷骰子 的 次数的期望: 第一个面第一次出现的概率是p1 n/n; 第二个面第一次出现的概率是p2 (n-1)/n; 第三个面第一次出现的概率是p3 (n-2)/n; ... 第 i 个面第一次出现的概率是pi (n-i+1)/n; 先看一下什么是几何分布: 几何分布: 在第n次伯努利试验中,试验 k 次才得到第一次成功…

题意:给出一个n面的色子,问看到每个面的投掷次数期望是多少. 析:这个题很水啊,就是他解释样例解释的太...我鄙视他,,,,, dp[i] 表示 已经看到 i 面的期望是多少,然后两种选择一种是看到新的一面,另一种是看到旧的一面,然后就很出答案了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include &…