http://codeforces.com/contest/888 A Local Extrema[水] [题意]:计算极值点个数 [分析]:除了第一个最后一个外,遇到极值点ans++,包括极大和极小 [代码]: #include<bits/stdc++.h> using namespace std; int main() { +]; int maxn,minn; maxn=minn=; cin>>n; ;i<=n;i++) { cin>>a[i]; } ;i&l…
You are given an array a consisting of n integers, and additionally an integer m. You have to choose some sequence of indices b1, b2, ..., bk (1 ≤ b1 < b2 < ... < bk ≤ n) in such a way that the value of is maximized. Chosen sequence can be empty.…
C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given n points on the straight line — the positions (x-coordinates) of the cities and m points on the same line…
题目链接 题意:给你两个数n,m,和一个大小为n的数组. 让你在数组找一些数使得这些数的和模m最大. 解法:考虑 dfs但是,数据范围不允许纯暴力,那考虑一下折半搜索,一个从头开始往中间搜,一个从后往中间搜.在中间相遇的时间二分更新最大值即可. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mp make_pair #define pb push_back usin…
1) Link to the problem: http://codeforces.com/contest/888/problem/C 2) Description: You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant if each substring of s with length at least k contains this character…
题目链接:E. Maximum Subsequence 用了一个Meet-in-the-middle的技巧,还是第一次用到这个技巧,其实这个技巧和二分很像,主要是在dfs中,如果数量减小一半可以节约很多的时间. Meet in the middle(有时候也叫作split and merge)是一种用以获取足够高效解决方案的灵巧的思想.和分治思想非常类似,它将问题分割成两个部分,然后试着合并这两个子问题的结果.好处在于通过使用一点额外的空间,你可以解决两倍规模的原来可以解决的问题. #in…
A permutation p of size n is an array such that every integer from 1 to n occurs exactly once in this array. Let's call a permutation an almost identity permutation iff there exist at least n - k indices i (1 ≤ i ≤ n) such that pi = i. Your task is t…
[Educational Codeforces Round 16]D. Two Arithmetic Progressions 试题描述 You are given two arithmetic progressions: a1k + b1 and a2l + b2. Find the number of integers x such that L ≤ x ≤ R andx = a1k' + b1 = a2l' + b2, for some integers k', l' ≥ 0. 输入 Th…
