题目链接:http://codeforces.com/contest/762/problem/D 多多分析状态:这个很明了 #include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define…

Educational Codeforces Round 17 A. k-th divisor 水题,把所有因子找出来排序然后找第\(k\)大 view code //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<bits/stdc++.h> using namespace std; function<voi…

D. Longest Subsequence 题目连接: http://www.codeforces.com/contest/632/problem/D Description You are given array a with n elements and the number m. Consider some subsequence of a and the value of least common multiple (LCM) of its elements. Denote LCM a…

A. Maximum Increase time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given array consisting of n integers. Your task is to find the maximum length of an increasing subarray of the gi…

https://codeforces.com/contest/1132/problem/F 思维 + 区间dp 题意 给一个长度为n的字符串(<=500),每次选择消去字符,连续相同的字符可以同时消去,问最少需要消去多少次 题解 定义dp[l][r]为区间[l,r]剩下一个字符所需要的最小次数 dp[l][r]=min(dp[l][i]+dp[i+1][r]+x) x为消去剩下两个字符所需要的次数,假如两个字符相同需要x=-1 代码 #include<bits/stdc++.h> #de…

https://codeforces.com/contest/1051/problem/D 题意 一个2*n的矩阵,你可以用黑白格子去填充他,求联通块数目等于k的方案数,答案%998244353. 思路 按列dp,定义状态dp[i][j][k]为前i列,有j个联通块,最后一列为k的方案数 处理出列状态之间的转移(会新产生多少个新的联通块) 注意初始化问题 #include<bits/stdc++.h> #define P 998244353 #define ll long long using…

题目链接:http://codeforces.com/contest/598/problem/E E. Chocolate Bar time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You have a rectangular chocolate bar consisting of n × m single squares. Y…

又被虐了... (记一次惨痛的Codeforces) 好不容易登上去了Codeforces,22:35准时开打 第一题,一看:这不SB题嘛?直接枚举因数上啊.9min才过掉了pretest 第二题... ... 什么鬼东西??? 写了个网络流上去 TLE 看来我还是太天真了 第三题呢? 貌似是个dp,不会... 剩下的没心情看了...然后老妈催我睡觉. 于是我一觉睡到了七八点 躺床上,我不甘心又去想B题,发现好像...直接贪心模拟可做!!! 立马爬了起来去码,测样例才发现刚刚想的做法没有考虑到每…

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…

C. Two strings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another)…