A. Adjacent Replacements time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mishka got an integer array aa of length nn as a birthday present (what a surprise!). Mishka doesn't like this prese…

http://codeforces.com/contest/1006/problem/A Mishka got an integer array aa of length nn as a birthday present (what a surprise!). Mishka doesn't like this present and wants to change it somehow. He has invented an algorithm and called it "Mishka's A…

                                                                           . A. Adjacent Replacements 执行1e9次命令,输出的最后数组的样子 一个奇数执行两次命令 会变回原来的数字 一个偶数只会执行一次命令 变成比自身小1的数 #include<bits/stdc++.h> #define int long long #define MAX(a,b,c) max(a,max(b,c)) #de…

[比赛链接] https://codeforces.com/contest/1006 [题解] Problem A. Adjacent Replacements        [算法] 将序列中的所有偶数替换为奇数即可 时间复杂度 : O(N) [代码] #include<bits/stdc++.h> using namespace std; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y);…

1.Universal selectors eg:#target*{ } 2.Child selectors < something immediately nested within something直接子女,不包括子孙 3.Adjacent Selector + something immediately following something 直接跟在后面的第一个跟屁虫 CSS3:h1 ~ p { font-weight: bold } will style all paragraphs…

[CC-ADJLEAF2]Adjacent Leaves 题目大意: 给定一棵有根树,考虑从根开始进行DFS,将所有叶子按照被遍历到的顺序排列得到一个序列. 定义一个叶子集合合法,当且仅当存在一种DFS的方式使得这个叶子集合在序列中的出现位置是一个连续子串. 给出一个\(n(n\le5\times10^5)\)个点的无根树,\(m(m\le5\times10^5)\)次询问,求以\(R\)为根的情况下叶子集合\(S(\sum|S|\le5\times10^5)\)是否合法. 思路: 定义\(le…

Adjacent Bit Counts 4557 Adjacent Bit CountsFor a string of n bits x 1 , x 2 , x 3 ,..., x n , the adjacent bit count of the string (AdjBC(x)) is given byx 1 ∗ x 2 + x 2 ∗ x 3 + x 3 ∗ x 4 + . . . + x n−1 ∗ x nwhich counts the number of times a 1 bit…

SSRS 报表中有一些高级的技巧,平常很少用到,下面我通过这个案例来展现一下如何在实际开发中使用它们,并且如何解决一些实际的需求. 这张报表分别统计了不同的 Product 产品在不同的月份的 Order 订单数量, Due 付款数量和 Ship 装船数量. Start Date 和 End Date 的时间范围作为筛选,很显然第一个 Matrix 是以 Order Date 作为比较条件,第二个是以 Due Date, 第三是以 Ship Date 作为比较条件. 现在需要变成这样的一种需求…

描述 For a string of n bits x1, x2, x3, …, xn,  the adjacent bit count of the string  is given by     fun(x) = x1*x2 + x2*x3 + x3*x 4 + … + xn-1*x n which counts the number of times a 1 bit is adjacent to another 1 bit. For example: Fun(011101101) = 3…

For a string of n bits x1, x2, x3,…, xn, the adjacent bit count of the string (AdjBC(x)) is given by x1 ∗ x2 + x2 ∗ x3 + x3 ∗ x4 + . . . + xn−1 ∗ xn which counts the number of times a 1 bit is adjacent to another 1 bit. For example: AdjBC(011101101)…