Pocky Let’s talking about something of eating a pocky. Here is a Decorer Pocky, with colorful decorative stripes in the coating, of length L. While the length of remaining pocky is longer than d, we perform the following procedure. We break the pocky…

对长为L的棒子随机取一点分割两部分,抛弃左边一部分,重复过程,直到长度小于d,问操作次数的期望. 区域赛的题,比较基础的概率论,我记得教材上有道很像的题,对1/len积分,$ln(L)-ln(d)+1$. /** @Date : 2017-10-06 14:32:03 * @FileName: HDU 5984 数学期望.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://gi…

题意是给定一长为 L 的木棒,每次任意切去一部分直到剩余部分的长度不超过 D,求切割次数的期望. 若木棒初始长度不超过 D,则期望是 0.000000: 设切割长度为 X 的木棒切割次数的期望是 F(X). 则 F(X) = F(切割点位置为 0 ~ D) + F(切割点位置为 D ~ X ) + 1:(此处的 +1 是指首次切割产生的次数) 而 F(切割点位置为 0 ~ D ) = 0:(因为已无需再切割) 令下一次切割点的位置为 T, F(切割点位置为 D ~ X ) = 在D~X上积分 (…

PockyTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2095 Accepted Submission(s): 1133 Problem DescriptionLet’s talking about something of eating a pocky. Here is a Decorer Pocky, with colorful dec…

Let’s talking about something of eating a pocky. Here is a Decorer Pocky, with colorful decorative stripes in the coating, of length L. While the length of remaining pocky is longer than d, we perform the following procedure. We break the pocky at an…

摘要 本文主要列举并求解了2016 ACM/ICPC亚洲区青岛站现场赛的部分真题,着重介绍了各个题目的解题思路,结合详细的AC代码,意在熟悉青岛赛区的出题策略,以备战2018青岛站现场赛. HDU 5984 Pocky 题意 给出一根棒子(可以吃的)的长度x和切割过程中不能小于的长度d,每次随机的选取一个位置切开,吃掉左边的一半,对右边的棒子同样操作,直至剩余的长度不大于d时停止.现在给出x和d,问切割次数的数学期望是多少. 解题思路 当看到第二个样例2 1时,结果是1.693147,联想到ln…

http://www.oyohyee.com/post/HDU/5984.html 看这篇吧,懒得写了. 训练时推得的式子有点鬼畜. #include<cstdio> #include<cmath> using namespace std; int n; double d,L; int main(){ // freopen("c.in","r",stdin); scanf("%d",&n); for(int i=1…

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Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…

http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…