C3 - Encryption (hard) 思路: 记sum[i]表示0 - i 的和对 p 取模的值. 1.如果k * p > n,那么与C2的做法一致,O(k*p*n)复杂度低于1e8. 2.如果k * p <= n 那么根据抽屉原理,必有k个sum[i]相同, 那么任意取k - 1个相同的 sum[i],记它们的下标为 l1,l2,......,lk-1 ,那么显然区间[li + 1, li+1](1<=i<k-1)的贡献为0 有贡献的区间只有[1,l1]和[lk-1 +…

转自:http://www.cnblogs.com/widsom/p/8863005.html 题目大意: 比起Encryption 中级版,把n的范围扩大到 500000,k,p范围都在100以内,然后让你求最小值 基本思路: 记sum[i]表示0 - i 的和对 p 取模的值. 1.如果k * p > n,那么与C2的做法一致,O(k*p*n)复杂度低于1e8. 2.如果k * p <= n 那么根据抽屉原理,必有至少k个sum[i]相同, 那么任意取k - 1个相同的 sum[i],记它…

大意: 给定序列$a$, 要求将$a$分成$k$个非空区间, 使得区间和模$p$的和最小, 要求输出最小值. $k$和$p$比较小, 直接暴力$dp$, 时间复杂度是$O(nklogp)$, 空间是$O(nk+kp)$ $dp[i][j]=min(...,f[j-1][s[i]-1]+1,f[j][s[i]],f[j][s[i]+1]-1+p,...)$ 看了其他提交, 好像有$O(nk)$的做法. #include <iostream> #include <sstream> #i…

B - Reversing Encryption A string s of length n can be encrypted by the following algorithm: iterate over all divisors of n in decreasing order (i.e. from n to 1), for each divisor d, reverse the substring s[1-d] (i.e. the substring which starts at p…

Hidden Code 题目连接: http://codeforces.com/gym/100015/attachments Description It's time to put your hacking skills to the test! You've been called upon to help crack enemy codes in the current war on... something or another. Anyway, the point is that yo…

A. Right-Left Cipher time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Polycarp loves ciphers. He has invented his own cipher called Right-Left. Right-Left cipher is used for strings. To encr…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Polycarp loves ciphers. He has invented his own cipher called Right-Left. Right-Left cipher is used for strings. To encrypt the string s=s1s2…

在CRM的日常开发中,Data Encryption经常是不得不开启的一个功能.但是有时,我们可能遇到一种情况,Organization导入之后,查看Data Encryption是已激活的状态,但是Key却是空的,导致我们不能激活.笔者碰到过类似的情况,当时用的解决方法是清空激活的Key,然后从新激活,具体的操作是在SQL中执行如下的语句: use ***_MSCRM update [EmailServerProfileBase] set IncomingPassword=null; upda…

上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题.. 今天,我们来扒一下cf的题面! PS:本代码不是我原创 1. 必要的分析 1.1 页面的获取 一般情况CF的每一个 contest 是这样的: 对应的URL是:http://codeforces.com/contest/xxx 还有一个Complete problemset页面,它是这样的:…

http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of bconsecutive cells. No cell can be part of two ships, however, the shi…