题意: Duff每天要吃ai千克肉,这天肉的价格为pi(这天可以买好多好多肉),现在给你一个数值n为Duff吃肉的天数,求出用最少的钱满足Duff的条件. 思路: 只要判断相邻两天中,今天的总花费 = ai*pi 与昨天的总花费(还有加上今天要吃的肉的重量)= (ai-1 + ai)*pi-1 . 代码如下: #include <iostream> #include <cstdio> #include <cstring> #include <fstream>…

题意: 一个数x被定义为lovely number需要满足这样的条件:不存在一个数a(a>1),使得a的完全平方是x的因子(即x % a2  != 0). 给你一个数n,求出n的因子中为lovely number的最大因子. 思路: 由于1<=n<=0=109,所以我们只要从1到找sqrt(n)就好,然后先从最大因子开始判断是否满足lovely number的条件,当满足时直接出输出即可. 代码如下: #include <iostream> #include <cstd…

D. Duff in Beach Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/588/problem/D Description While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. Th…

C. Duff and Weight Lifting Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/588/problem/C Description Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of wei…

B. Duff in Love Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/588/problem/B Description Duff is in love with lovely numbers! A positive integer x is called lovely if and only if there is no such positive integer a > 1 such…

A. Duff and Meat Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/588/problem/A Description Duff is addicted to meat! Malek wants to keep her happy for n days. In order to be happy in i-th day, she needs to eat exactly aikilo…

B. Pasha and PhonePasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits. Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a1, …

题意(CodeForces 588E) 给定一棵\(n\)个点的树,给定\(m\)个人(\(m\le n\))在哪个点上的信息,每个点可以有任意个人:然后给\(q\)个询问,每次问\(u\)到\(v\)上的路径有的点上编号最小的\(k(k \le 10)\)个人(没有那么多人就该有多少人输出多少人). 分析 \(u\)到\(v\)上路径的询问很显然的想到LCA,但是要维护前\(k\)个在路径上的最小的点似乎是个有点麻烦的问题.其实,找到了LCA(设为\(p\)点),我们就可以同样的利用倍增的思想…

B. Duff in Love time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Duff is in love with lovely numbers! A positive integer x is called lovely if and only if there is no such positive integer…

题意:一个n个点的数, m个人住在其中的某些点上, 每个人的标号1-m, 询问u-v 路径上标号前a个人,并输出标号,a < 10. 作法, 利用倍增, ID[j][i] 表示i到i的第2^j个祖先上前10个人, 那么每次询问直接维护就好了,细节好多, 刚开始不知道怎么求ID[j][i]. 这里把2^j分成两部分, 前2^(j-1)和 后2^(j-1)个, 然后递推的维护. 感觉树链剖分也可以做, 不知道会不会TLE, 树链剖分的话 线段树的每个点维护10个值, 每次合并就行了. #includ…