Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <xx, yy>. If x_ixi​ = x_jxj​and y_iy…

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx , 00 ), ( 00 , yy ), ( xx , yy ). Every time the wave will wash out the trace of f…

Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That means, if he reads books from ll to rr, he will get a…

Trace There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xxx , yyy ) means the wave is a rectangle whose vertexes are ( 000 , 000 ), ( xxx , 000 ), ( 000 , yyy ), ( xxx , yyy ). Every time the wave will wash o…

ACM-ICPC 2018 徐州赛区网络赛  去年博客记录过这场比赛经历:该死的水题  一年过去了,不被水题卡了,但难题也没多做几道.水平微微有点长进.     D. Easy Math 题意:   给定 \(n\), \(m\) ,求 \(\sum _{i=1}^{m} \mu(in)\) .其中 $ 1 \le n \le 1e12$ , $ 1 \le m \le 2e9$ ,\(\mu(n)\) 为莫比乌斯函数.   思路:   容易知道,\(i\) 与 \(n\) 不互质时, \(\m…

目录 A. Hard to prepare B.BE, GE or NE F.Features Track G.Trace H.Ryuji doesn't want to study I.Characters with Hash J. Maze Designer A. Hard to prepare After Incident, a feast is usually held in Hakurei Shrine. This time Reimu asked Kokoro to deliver…

A prime number (or a prime) is a natural number greater than 11 that cannot be formed by multiplying two smaller natural numbers. Now lets define a number NN as the supreme number if and only if each number made up of an non-empty subsequence of all…

题库链接: https://nanti.jisuanke.com/t/41387 题目大意 给定n个数,与一个数m,求ai右边最后一个至少比ai大m的数与这个数之间有多少个数 思路 对于每一个数,利用二分的方法求他右边大于等于ai+m的数的最后一个值. 关键在于怎么二分呢? 利用线段树存储区间最大值,看这个区间的最大值是不是比ai+m大 代码: #include<bits/stdc++.h> using namespace std; #define maxn 1000005 #define m…

262144K   Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <x, y>. If xi​= xj​ and yi…

ACM-ICPC 2018 徐州赛区网络预赛 G. Trace (思维,贪心) Trace 问答问题反馈 只看题面 35.78% 1000ms 262144K There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx ,…

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xxx , yyy ) means the wave is a rectangle whose vertexes are ( 000 , 000 ), ( xxx , 000 ), ( 000 , yyy ), ( xxx , yyy ). Every time the wave will wash out the…

ACM-ICPC 2018 徐州赛区网络预赛 A.Hard to prepare 枚举第一个选的,接下来的那个不能取前一个的取反 \(DP[i][0]\)表示选和第一个相同的 \(DP[i][1]\)表示选和第一个取反的 \(DP[i][2]\)表示选其他的 状态转移方程直接看代码好了 //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #inc…

H.Ryuji doesn't want to study 27.34% 1000ms 262144K   Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That…

传送门 题面: In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to consta…

In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to constantly fac…

262144K   Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That means, if he reads books from ll to rr, he…

Mur loves hash algorithm, and he sometimes encrypt another one's name, and call him with that encrypted value. For instance, he calls Kimura KMR, and calls Suzuki YJSNPI. One day he read a book about SHA-256,which can transit a string into just 256 b…

链接 https://nanti.jisuanke.com/t/31456 参考题解  https://blog.csdn.net/ftx456789/article/details/82590044 #include <bits/stdc++.h> #define pb push_back #define mp make_pair #define fi first #define se second #define all(a) (a).begin(), (a).end() #define…

ACM-ICPC 2018 徐州赛区网络预赛 J. Maze Designer J. Maze Designer After the long vacation, the maze designer master has to do his job. A tour company gives him a map which is a rectangle. The map consists of N \times MN×M little squares. That is to say, the h…

F. Features Track Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <xx, yy>. If x_ixi…

A. Hard to prepare 题意:有n个客人做成一圈,有$2^k$种面具,对于每种面具有一种面具不能使相邻的两个人戴,共有多少种做法. 思路: 把题意转化成相邻的人不能带同种面具.总数为$(2^k)^n$,减去一对相邻的客人戴同种面具$(2^k)^{(n-1)}*C(n,1)$,其中重复了两对相邻的客人戴同种面具$(2^k)^{(n-2)}*C(n,2)$,依次容斥. 最后所有人都戴同种面具的情况额外考虑,当n是奇数时,n-1对客人相同即所有人相同.n为偶数时,n-1对客人相同时用公式…

历程:由于只是网络赛,所以今天就三开了.一开始的看题我看了d题,zz和jsw从头尾看起来,发现c题似乎可做,和费马大定理有关,于是和zz一起马上找如何计算勾股数的方法,比较慢的A掉了,而jsw此时看了最后的两道题,并且也与此同时把倒数第二题做掉了,但是此时hdu评测机炸了,判题队列排到几十页之后,所以不知道是否ac,然后就开始想最后一道题目,我和zz想了一下以为是一道类似dp或者图论题目,而jsw想到了线段树的做法,我们稍微讨论了一下就得到了正解,但是由于三开,而我每次想代码的时候都要把所有细节…

A. Hard to prepare #include <bits/stdc++.h> using namespace std; ; ]; ]; int main() { int T; int n,k; bit[] = ; ; i <= ; i++) { bit[i] = *bit[i-]%MOD; } scanf("%d", &T); while (T--) { scanf("%d%d", &n, &k); f[] = b…

D.Made In Heaven One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. However, Pucci the father somehow knows it and wants to stop her. There are NN spots in the jail and MM roads connecting some of the spots. JOJO fi…

这次网络赛没有打.生病了去医院了..尴尬.晚上回来才看了题补简单题. K  Supreme Number 题目链接:https://nanti.jisuanke.com/t/31452 题意:输入一个整数n(其实可以当成字符串,因为很长啊.),求满足不超过n的supreme number.这个supreme number的定义是,这个字符串的子串都是质数.比如.137就是,但是19就不是. 题解:找规律.我先开始题没看懂,没懂子串也要为素数.后面才看到.然后在纸上枚举了一下可能出现的情况.发现大…

https://nanti.jisuanke.com/t/31458 题意 有N个帧,每帧有K个动作特征,每个特征用一个向量表示(x,y).两个特征相同当且仅当他们在不同的帧中出现且向量的两个分量分别相等.求最多连续相同特征的个数? 分析 用一个map来维护帧中特征的信息,map中的键即读入的向量,因此用一个pair<int , int>表示. 键的值也是一个pair,需要记录它当前已经连续了多少次,还需要记录它上一次出现的帧的位置.如果它上一帧没有出现过,那么它的连续次数就要被置成1:如果上…

签到题 因为一个小细节考虑不到wa了两次 // 一开始没这个if wa了.因为数据中存在同一帧(frame)一个相同的值出现多次,这样子同一个i 后面的同样的特征会把len重置为1 #include <bits/stdc++.h> using namespace std; typedef long long ll; int t; int n; struct val { int last; int len; void update(int t) { if (t == last + 1) ++le…

https://nanti.jisuanke.com/t/31459 题意 n个矩阵,不存在包含,矩阵左下角都在(0,0),给右上角坐标,后来的矩阵会覆盖前面的矩阵,求矩阵周长. 分析 set按照x或者y从大到小排序,从后往前遍历,放入set,找到当前矩阵在set中的位置,当前矩阵的x或者y,与set中后一位矩阵的x或者y的差值,就是增加的横线或者竖线的长度 感觉套个线段树求周长的模板也行. #include<queue> #include<cstring> #include<…

题目链接:https://codeforces.com/gym/102012/problem/I 题意:问有多少个 1 到 n 的排列,使得用给定的 k 个比较器(使 au 和 av 有序)排序后,整个序列的最长上升子序列为 n - 1. 题解:先处理出全部最长上升子序列为 n - 1 的排列,然后枚举每个比较器使用与否,计数即可.(题目卡常,dfs要加引用...) #include <bits/stdc++.h> using namespace std; #define ll long lo…

呃.自闭了自闭了.我才不会说我写D写到昏天黑地呢. I  Characters with Hash 题目链接:https://nanti.jisuanke.com/t/31461 题意:给你一个字符串s,一个种子字母L.根据 |int(L) - s[i]|公式的到hash后的字符,如果是个位数就变成两位数,最后去掉前导0计算字符串长度.也就是0->00 7->07.最后统计如果前面有0就都去掉.如果在中间qwq就不用啦. 题解:暴力模拟//先开始卡题意..初始长度就设成2n好了.如果判断前面有…