设dp_i为已经出现了i面,需要的期望次数,dp_n=0 那么dp_i= i/n*dp_i + (n-i)/n*dp_(i+1) + 1 现在已经i面了,i/n的概率再选择一次i面,(n-i)/n的概率选到新的一面,分别乘其期望次数,并且这次丢过,所以+1 #include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; typedef pair<int,…

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…

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…

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…

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…

Brush (III) LightOJ - 1017 题意:有一些点,每刷一次可以将纵坐标在区间(y1,y1+w)范围内的所有点刷光,y1为任何实数.最多能刷k次,求最多共能刷掉几个点. 先将点按照纵坐标从小到大排序. 显然,横坐标没有任何作用.记p[i]为排序后第i个点的纵坐标. 显然,每一次以某个点的纵坐标为y1来刷,一定不会比以其他的数为y1来刷更差. 记x[i]为以第i个的纵坐标为y1来刷能刷掉的点的数量.容易预处理出来. ans[i][j]表示以第i个的纵坐标为y1,刷j次能刷掉的点数…

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…

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…

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 &…