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Random Access Iterator \[ Time Limit: 4000 ms \quad Memory Limit: 262144 kB \] 题意 给出伪代码,问按着伪代码在树上跑,能够正确求出来树的深度的概率. 思路 先在树上 \(dfs\) 一遍,求出每个点可以走到的最深深度,用 \(deep\) 表示. 令 \(dp[i]\) 表示从 \(i\) 节点能够正确求出最深深度的概率. 如果 \(deep[u]\) 不等于最深深度,那么明显 \(dp[u]=0\) 如果 \(de…
以太坊MPT树的持久化层是采用了leveldb数据库,然而在抽取MPT树代码运行过程中,进行get和write操作时却发生了错误: Caused by: org.fusesource.leveldbjni.internal.NativeDB$DBException: IO error: C:\data\trie\.sst: Could not create random access file. at org.fusesource.leveldbjni.internal.NativeDB.che…
Random Access Iterator 树型概率DP dp[u]代表以当前点作为根得到正确结果的概率 将深度最深的几个点dp[u]很明显是1 然后很简单的转移 有k次,但我们要先看一次的情况,然后再推到k次,k次中只要有一次就可以正确,所以求出k次全失败的概率,用1去减即可 #include <bits/stdc++.h> #define ll long long using namespace std; const int maxn = 1e6+7; const int mod = 1…
问题描述:        把列表listview写入notebook里. 在main函数中, win = create_and_set_a_window(); book = gtk_notebook_new(); gtk_notebook_set_tab_pos( GTK_NOTEBOOK( book ), GTK_POS_LEFT ); gtk_container_add( GTK_CONTAINER( win ), book ); create_page_one( book, win );…
题目链接 大致题意:从根节点出发,在节点x有son[x]次等概率进入儿子节点,求到达最深深度的概率.son[x]为x节点的儿子节点个数. 又又又又没做出来,心态崩了. 下来看了官方题解后发觉自己大体思路是没错的,但是细节太弱了Orz. 大体思路:设dp[x]为以x为根节点,求到达最深深度的概率.先跑一遍dfs,求出每个点的子节点数,每个点的深度以及最深深度. 我们可以求得从x点一次不能到达最深深度的概率为$cnt =1-(\tfrac{\sum_{y\epsilon x } dp[y]}{son…
每次循环向下寻找孩子时,随机选取一个孩子,设dp[u]为从u出发,不能得出正确答案的概率,则从u出发,走一次的情况下不能得出正确答案的概率是 P = (dp[v1]+dp[v2]+dp[v3]+--dp[vk]) / cnt_son[u] ,则从u出发,要走cnt_son[u]次,那么dp[u]=P^cnt_con[u] dp的意义也可以改成能得出正确答案的概率,下面的式子稍微改改就行了 为了避免除法的精度问题,num/k %mod,它等于 num * ni %mod ,ni等于k在模mod意义…
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文章首发于浩瀚先森博客http://www.guohao1206.com/2016/12/07/1248.html 说到计算机上的存储器,很多人第一反应是硬盘,然后是内存. 其实在计算机上除了硬盘和内存属于存储器以外CPU寄存器和Cache他们也是属于一类的. 对于存储器我们在乎的无外乎两个属性----"快"和"大",我们希望存储器容量越大越好,传输速度越快越好,然后鱼和熊掌不能兼得.由于现在硬件技术的限制,我们能制造错容量虽小但是速度很快的存储器,也可以制造出容量…
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