既然补了就简单记录一下. 感觉还算有一点营养. 官方题解传送门:点我 A Commentary Boxes 对拆掉$n \mod m$个和新建$m - (n \mod m)$求个最小. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; int main() { ll n, m, a, b, r; scanf("…

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…

  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<…

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<…

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 "+"…

A. 一个小模拟    不解释 //By SiriusRen #include <bits/stdc++.h> using namespace std; long long n,m,a,b; int main(){ scanf("%I64d%I64d%I64d%I64d",&n,&m,&a,&b); printf("%I64d\n",min((n%m)*b,(m-(n%m))*a)); } B. 给你一个数列A[i] 把所…

[Educational Codeforces Round 16]E. Generate a String 试题描述 zscoder wants to generate an input file for some programming competition problem. His input is a string consisting of n letters 'a'. He is too lazy to write a generator so he will manually ge…

[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…