[题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=4397 [算法] 树状数组 时间复杂度 : O(QlogN) [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 100010 int n , q; struct Binary_Indexed_Tree { int c[MAXN]; inline int lowbit(int x) { return x &a…

4397: [Usaco2015 dec]Breed Counting Time Limit: 10 Sec  Memory Limit: 128 MB Description Farmer John's N cows, conveniently numbered 1…N, are all standing in a row (they seem to do so often that it now takes very little prompting from Farmer John to…

4397: [Usaco2015 dec]Breed Counting Time Limit: 10 Sec  Memory Limit: 128 MB Submit: 29  Solved: 25 [Submit][Status][Discuss] Description Farmer John's N cows, conveniently numbered 1-N, are all standing in a row (they seem to do so often that it now…

bzoj4397[Usaco2015 dec]Breed Counting 题意: 给定一个长度为N的序列,每个位置上的数只可能是1,2,3中的一种.有Q次询问,每次给定两个数a,b,请分别输出区间[a,b]里数字1,2,3的个数.n≤100000,q≤100000. 题解: 裸前缀和. 代码: #include <cstdio> #include <cstring> #include <algorithm> #define inc(i,j,k) for(int i=j…

题目描述 Farmer John's N cows, conveniently numbered 1…N, are all standing in a row (they seem to do so often that it now takes very little prompting from Farmer John to line them up). Each cow has a breed ID: 1 for Holsteins, 2 for Guernseys, and 3 for…

原题链接https://www.lydsy.com/JudgeOnline/problem.php?id=4397 用线段树维护区间和即可.时间复杂度\(O((N+Q)logN)\). #include<iostream> #include<cstring> #include<cstdio> #define maxn 100010 using namespace std; inline int read(){ register int x(0),f(1); regist…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

本篇博客按照题号排序(带*为推荐题目) 1008 [HNOI2008]越狱 很经典的题了..龟速乘,龟速幂裸题,, 1010 [HNOI2008]玩具装箱toy* 斜率优化 基本算是裸题. 1012 最大数 单调队列/单调栈 随便搞一搞就好 (水题...) 1045 糖果传递(双倍经验请做 3293 分金币) 贪心+中位数  排序后搞一搞可过..(水题...) 1051 受欢迎的牛* tarjan (提高T1+ -- T2难度) 1106: [POI2007]立方体大作战tet* 树状数组+贪心…

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…