Problem Figure 2. The Hamming distance between these two strings is 7. Mismatched symbols are colored red. Given two strings ss and tt of equal length, the Hamming distance between ss and tt, denoted dH(s,t)dH(s,t), is the number of corresponding sym…

Problem Figure 2. The Hamming distance between these two strings is 7. Mismatched symbols are colored red. Given two strings ss and tt of equal length, the Hamming distance between ss and tt, denoted dH(s,t)dH(s,t), is the number of corresponding sym…

在上篇,我了解了基数的基本概念,现在进入Linear Counting算法的学习. 理解颇浅,还请大神指点! http://blog.codinglabs.org/articles/algorithms-for-cardinality-estimation-part-ii.html 它的基本处理方法和上篇中用bitmap统计的方法类似,但是最后要用到一个公式: 说明:m为bitmap总位数,u为0的个数,最后的结果为n的一个估计,且为最大似然估计(MLE). 那么问题来了,最大似然估计是什么东东…

来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31474   Accepted: 15724 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is repr…

ZOJ3944 People Counting ZOJ3939 The Lucky Week 1.PeopleConting 题意:照片上有很多个人,用矩阵里的字符表示.一个人如下: .O. /|\ (.) 占3*3格子,句号“.”为背景.没有两个人完全重合.有的人被挡住了一部分.问照片上有几个人. 题解: 先弄个常量把3*3人形存起来,然后6个部位依次找,比如现在找头,找到一个头,就把这个人删掉(找这个人的各个部位,如果在该部位位置的不是这个人的身体,就不删),删成句号,疯狂找就行了. 代码:…

#include <stdio.h> #include <malloc.h> #define MAX_STACK 10 ; // define the node of stack typedef struct node { int data; node *next; }*Snode; // define the stack typedef struct Stack{ Snode top; Snode bottom; }*NewStack; void creatStack(NewSt…

1004. Counting Leaves (30)   A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100,…

Problem A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains. An example of a length 21 DNA string (whose alphabet contains the symbol…

今天在学习vuex的过程中,遇到了一个很困扰人的问题,最终利用vuex的状态快照工具logger解决了问题. 问题是这样的,我在子组件中使用了mapState()函数来将状态映射至子组件中,使子组件能够获取到状态,但是发现除了第一次能够得到状态的初始值外,后面无论状态如何变更,都获取不到更新的结果. 这是子组件mapState部分的代码: computed: mapState({ user: ({user}) => (user), session: ({sessions, currentSess…

// uva 11401 Triangle Counting // // 题目大意: // // 求n范围内,任意选三个不同的数,能组成三角形的个数 // // 解题方法: // // 我们设三角巷的最长的长度是c(x),另外两边为y,z // 则由z + y > x得, x - y < z < x 当y = 1时,无解 // 当y = 2时,一个解,这样到y = x - 1 时 有 x - 2个 // 解,所以一共是0,1,2,3....x - 2,一共(x - 2) * (x - 1…