目录 Codeforces 990 A.Commentary Boxes B.Micro-World C.Bracket Sequences Concatenation Problem D.Graph And Its Complement(思路 构造) E.Post Lamps(贪心) F.Flow Control(思路) G.GCD Counting(思路) Codeforces 990 比赛链接 真特么菜啊 后一个多小时无所事事.. 要多做CF的题啊,也许没有那么难但是就是容易想不到,代码也…

目录 Codeforces 1009 A.Game Shopping B.Minimum Ternary String C.Annoying Present D.Relatively Prime Graph E.Intercity Travelling(递推) \(Description\) \(Solution\) F.Dominant Indices(启发式合并) G.Allowed Letters(Hall定理 位运算) \(Description\) \(Solution\) Codef…

目录 Codeforces 1000 A.Codehorses T-shirts B.Light It Up C.Covered Points Count(差分) D.Yet Another Problem On a Subsequence(DP) E.We Need More Bosses(圆方树) \(Description\) \(Solution\) F.One Occurrence(线段树) \(Description\) \(Solution\) G.Two-Paths(树形DP)…

Educational Codeforces Round 84 (Div. 2) 读题读题读题+脑筋急转弯 = =. A. Sum of Odd Integers 奇奇为奇,奇偶为偶,所以n,k奇偶性要相同. 由求和公式得k个不同奇数组成的最小数为k2,所以n≥k2. #include <bits/stdc++.h> using namespace std; void solve(){ int n,k; cin>>n>>k; if((n-k)%2==0&&…

G - GCD Counting 思路:我猜测了一下gcd的个数不会很多,然后我就用dfs回溯的时候用map暴力合并就好啦. 终判被卡了MLE.....  需要每次清空一下子树的map... #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define pii pair<int,int> #define piii pair<…

  C. Bracket Sequences Concatenation Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A bracket sequence is a string containing only characters "(" and ")". A regular…

E - Post Lamps 思路:一开始看错题,以为一个地方不能重复覆盖,我一想值这不是sb题吗,直接每个power check一下就好....复杂度nlogn 然后发现不是,这样的话,对于每个power,假如我们覆盖到了x,那么我们要找到一个最大的 p <= x 且p 可以放灯,那么转移到的 为止为p + power,这样的话我想复杂度就变成了不是严格的nlogn,但是我写了一发还是过了,我感觉是复杂度接近nlogn,感觉没有 数据能把每个power的check都卡成n. #include<…

F - Flow Control 给你一个有向图,要求你给每条边设置流量,使得所有点的流量符合题目给出的要求. 思路:只有在所有点的流量和为0时有解,因为增加一条边的值不会改变所有点的总流量和, 所以我们dfs回溯的时候构造就好了, 其他边设为0. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define pii pair<int…

A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+"…

明明多个几秒就能场上AK了.自闭. A:签到. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long ')) c=getchar();return c;} ?n:gcd(m,n%m);} in…