只做了两个就去上课去啦... A. Company Merging time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output A conglomerate consists of nn companies. To make managing easier, their owners have decided to merge all co…

前言 有一场下午的cf,很滋磁啊,然后又和dalao(见右面链接)组队打了,dalao直接带飞我啊. 这是一篇题解,也是一篇总结,当然,让我把所有的题目都写个题解是不可能的了. 按照开题顺序讲吧. 在开始前有现场赛的成绩,所以可以看出来哪道是傻逼题,当然还是滋磁啊. M - The Pleasant Walk 我被分到了这道题,当然是因为我太弱了啊,dalao当然是要去做神仙题了. 好像没什么可说的了,直接扫不就行了.. #include<cstdio> #include<cstring…

2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred) easy: ACEGHK medium-easy: BJL medium: D ?????: I A. B. C. 对 \(R[],C[]\) 分别按奇偶性分段. 网! D. 考虑 check 一个串,枚举右走对应的前缀 pre,下走对应的后缀 suf. 把每行反串拼接中间连特殊字符,建 SA,能 match 上 p…

A. Berstagram Polycarp recently signed up to a new social network Berstagram. He immediately published n posts there. He assigned numbers from 1 to n to all posts and published them one by one. So, just after publishing Polycarp's news feed contained…

http://codeforces.com/contest/847/problem/D 巧妙的贪心 仔细琢磨... 像凸包里的处理 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <time.h> #include <string> #include <set> #include <map> #i…

A 思路: 贪心,每次要么选两个最大的,要么选三个,因为一个数(除了1)都可以拆成2和3相加,直到所有的数都相同就停止,这时就可以得到答案了; C: 二分+bfs,二分答案,然后bfs找出距离小于等于当前要判断距离的点,把这些点标记后,再遍历按价格排好序的商店,从小到大选择价格低的, 根据最后的数目和价格判断当前这个距离行不行,这样就可以得到答案了; G: 思路: 模拟,先判断是否可以放在要求的地方,不行的话就从前往后找能放的地方,这里可以把前面的区间[l,r]的右端点r+1当做当前区间的左端点…

I. Sale in GameStore(贪心) time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output A well-known Berland online games store has announced a great sale! Buy any game today, and you can download more g…

A. Toda 2 题意:给你n个人,每个人的分数是a[i],每次可以从两个人到五个人的使得分数减一,使得最终的分数相等: 思路:假设答案为m:每个人的分数与答案m的差值为d[i],sum为d[i]的总和,max为d[i]的最大值:仅当sum-max>=max的时候才满足: 满足之后,总和为奇数,先取三个,再取两个(都是最大与次大值):偶数每次取两个即可: B. Minimum and Maximum 题意:人机交互题:给你一个数组 ,找出其中的最小值与最大值,需要在询问f(n)的次数内得到最大…

J. Bottles time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottl…

G. Car Repair Shop time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Polycarp starts his own business. Tomorrow will be the first working day of his car repair shop. For now the car repair s…

题目链接:http://codeforces.com/problemset/problem/847/I I. Noise Level time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The Berland's capital has the form of a rectangle with sizes n × m quarte…

A. Find a Number 找到一个树,可以被d整除,且数字和为s 记忆化搜索 static class S{ int mod,s; String str; public S(int mod, int s, String str) { this.mod = mod; this.s = s; this.str = str; } } public static void main(String[] args) { IO io = new IO(); int[][]vis=new int[550…

A. Find a Number Solved By 2017212212083 题意:$找一个最小的n使得n % d == 0 并且 n 的每一位数字加起来之和为s$ 思路: 定义一个二元组$<d, s>$ 表示当前状态模d的值,以及每一位加起来的值 跑最短路,从$<0, 0>  跑到 <0, s>$ #include<bits/stdc++.h> using namespace std; ; const int INF = 0x3f3f3f3f; #de…

第一次打ACM比赛,和yyf两个人一起搞事情 感觉被两个学长队暴打的好惨啊 然后我一直做傻子题,yyf一直在切神仙题 然后放一波题解(部分) A. Find a Number LINK 题目大意 给你d和s,求你一个最小的数满足是d的倍数且数字和是s 思路 从高位到低位考虑广搜,把当前的长度和模d的余数作为状态,然后添加一个数就在对应的位置上加 一个模数只记录长度最小的状态,然后可以反着贪心回去 yyf的神仙code: #include <iostream> #include <cstd…

先把代码扔上来 E. Field of Wonders time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Polycarpus takes part in the "Field of Wonders" TV show. The participants of the show have to guess a hidde…

秉承ACM团队合作的思想懒,这篇blog只有部分题解,剩余的请前往星感大神Star_Feel的blog食用(表示男神汉克斯更懒不屑于写我们分别代写了下...) C. Cloud Computing 扫描线搞一搞区间(主席树也OK啊,只是空间玄学,主席树理论空间nlogn实际上开小那么10倍8倍没什么锅啊zzzz),对于权值建立权值线段树,然后记录每个权值出现的次数以及区间权值和,然后在线段树上二分求答案即. #include<cstdio> #include<cstring> #i…

题目链接:https://codeforces.com/contest/1090/problem/A A conglomerate consists of n companies. To make managing easier, their owners have decided to merge all companies into one. By law, it is only possible to merge two companies, so the owners plan to s…

题目链接:https://codeforces.com/contest/1090/problem/B Examplesstandard input The most famous characters of Pushkin’s works are Onegin \cite{onegin}, Dubrovsky \cite{dubrovsky} and Tsar Saltan \cite{saltan}. \begin{thebibliography}{99} \bibitem{saltan} A…

题目链接:https://codeforces.com/contest/1090/problem/D Vasya had an array of n integers, each element of the array was from 1 to n. He chose m pairs of different positions and wrote them down to a sheet of paper. Then Vasya compared the elements at these…

题目链接:https://codeforces.com/contest/1090/problem/M There are n houses along the road where Anya lives, each one is painted in one of k possible colors. Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have…

Tree Ming and Hong are playing a simple game called nim game. They have nn piles of stones numbered 11 to nn ,the ii-th pile of stones has a_iai​ stones. There are n - 1n−1 bidirectional roads in total. For any two piles, there is a unique path from…

感想: 今天三个人的状态比昨天计院校赛的状态要好很多,然而三个人都慢热体质导致签到题wa了很多发.最后虽然跟大家题数一样(6题),然而输在罚时. 只能说,水题还是刷得少,看到签到都没灵感实在不应该. 题目链接:http://acm.zju.edu.cn/onlinejudge/showContestProblems.do?contestId=391 A:简单贪心,按高度sort一下就好了,这里用优先队列处理 #include <cstdio> #include <queue> #i…

A. Company Merging Solved. 温暖的签到. #include<bits/stdc++.h> using namespace std; ; typedef long long ll; struct node{ int val, num; node(){} node(int val, int num):val(val), num(num){} }arr[maxn]; int n, m; int main() { while(~scanf("%d", &a…

Travel There are nn planets in the MOT galaxy, and each planet has a unique number from 1 \sim n1∼n. Each planet is connected to other planets through some transmission channels. There are mm transmission channels in the galaxy. Each transmission cha…

Swap There is a sequence of numbers of length nn, and each number in the sequence is different. There are two operations: Swap the first half and the last half of the sequence (if nn is odd, the middle number does not change) Swap all the numbers in…

Angel's Journey “Miyane!” This day Hana asks Miyako for help again. Hana plays the part of angel on the stage show of the cultural festival, and she is going to look for her human friend, Hinata. So she must find the shortest path to Hinata’s house.…

Tasks It's too late now, but you still have too much work to do. There are nn tasks on your list. The ii-th task costs you t_iti​seconds. You want to go to bed TT seconds later. During the TT seconds, you can choose some tasks to do in order to finis…

链接:https://nanti.jisuanke.com/t/39277 思路: 一开始看着很像树分治,就用树分治写了下,发现因为异或操作的特殊性,我们是可以优化树分治中的容斥操作的,不合理的情况只有当两点在一条链上才存在,那么直接一遍dfs从根节点向下跑途中维护一下前缀和,把所有情况中不合理情况造成的值修正. 这样的话时间复杂度就可以降得非常低了,感觉还可以优化,但是懒得写了 代码耗时:142ms. 实现代码: #include<bits/stdc++.h> using namespace…

链接:https://codeforc.es/gym/102267 A. Picky Eater 直接比较 int main(){ int x ,y; scanf("%d %d" ,&x ,&y); if(x>=y){ ; } ; ; } B. Primes 素数筛,log判断 ; int vis[maxn]; void is_prime(int N){ ;i<=N;i++){ if(!vis[i]){ prime[num_prime++] = i; vis…

题意:有个长度为\(n\)的监狱,犯人在位置\(a\),cop在位置\(b\),你每次可以向左或者向右移动一个单位,或者选择不动并在原地放一个爆竹\(i\),爆竹\(i\)在\(s[i]\)秒后爆炸,cop每次向你的位置移动一个单位,你最终一定会被抓住(因为监狱是有限的),问你在被抓住前,最多能看到多少爆竹爆炸. 题解:我们可以很容易算出最能放多少个爆竹不被cop抓住,和最晚被cop抓住的时间,然后对爆竹的爆炸时间排序,假如我们要放\(x\)个爆竹,那么很显然,最优的放法一定是在第一秒放\(s[…