"Oh, There is a bipartite graph.""Make it Fantastic."X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up seve…

正解: #include <bits/stdc++.h> using namespace std; const int INF = 0x3f3f3f3f; const int MAXN=10010;//点数的最大值 const int MAXM=400010;//边数的最大值 #define captype int struct SAP_MaxFlow{ struct EDGE{ int to,next; captype cap; }edg[MAXM]; int eid,head[MAXN];…

"Oh, There is a bipartite graph.""Make it Fantastic." X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up sev…

"Oh, There is a bipartite graph.""Make it Fantastic." X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up sev…

https://nanti.jisuanke.com/t/31447 题意 一个二分图,左边N个点,右边M个点,中间K条边,问你是否可以删掉边使得所有点的度数在[L,R]之间 分析 最大流不太会.. 贪心做法: 考虑两个集合A和B,A为L<=d[i]<=R,B为d[i]>R 枚举每个边 1.如果u和v都在B集合,直接删掉2.如果u和v都在A集合,无所谓3.如果u在B,v在A,并且v可删边即d[v]>L4.如果u在A,v在B,并且u可删边即d[u]>L 最后枚举N+M个点判断是…

关于有源上下界最大流: https://blog.csdn.net/regina8023/article/details/45815023 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n, m, k, l, r, s, t, superS, superT; ;//点数的最大值 ;//边数的最大值 const int INF = 0x3f3f3f3f; st…

2018 ICPC 沈阳网络赛 Call of Accepted 题目描述:求一个算式的最大值与最小值. solution 按普通算式计算方法做,只不过要同时记住最大值和最小值而已. Convex Hull 题目描述:定义函数\(gay(x)\),若\(x\)是某个非\(1\)的数的平方的倍数,则\(gay(x)=0\),否则\(gay(x)=x^2\),求\(\sum_{num=1}^{n} ( \sum_{i=1}^{num} gay(x) ) mod p\) solution \[\sum…

ACM-ICPC 南昌网络赛F题 Megumi With String 题目描述 给一个长度为\(l\)的字符串\(S\),和关于\(x\)的\(k\)次多项式\(G[x]\).当一个字符串\(str\)是S的子串时,定义\(str\)的\(value\)值为\(G[length(str)]\),否则\(value=0\).一个字符串的\(power\)值定义为其所有子串的\(value\)之和.每次向\(S\)末尾添加一个字符,重复\(m\)次.求每次添加字符后,一个长度为\(n\)的随机串\…

"Oh, There is a bipartite graph.""Make it Fantastic." X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up sev…

题目链接:https://nanti.jisuanke.com/t/31447 "Oh, There is a bipartite graph.""Make it Fantastic." X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between…

题目链接:哈哈哈哈哈哈 _(:з」∠)_ _(:з」∠)_ _(:з」∠)_ _(:з」∠)_ _(:з」∠)_ 哈哈哈哈哈哈,从9月16日打了这个题之后就一直在补这道题,今天终于a了,哈哈哈哈哈哈. 先把代码贴上,有时间再好好写题解,哈哈哈哈哈哈. ヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞ 代码,嘻嘻: #include<bits/stdc++.h> using namespace std; typedef long long ll…

哈哈哈哈哈哈哈哈哈哈哈哈,终于把这道题补出来了_(:з」∠)_ 来写题解啦. _(:з」∠)_ _(:з」∠)_ _(:з」∠)_ _(:з」∠)_ _(:з」∠)_ 哈哈哈哈哈哈,从9月16日打了这个题之后就一直在补这道题,今天终于a了,哈哈哈哈哈哈. 先把代码贴上,有时间再好好写题解,哈哈哈哈哈哈.ヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞヾ(◍°∇°◍)ﾉﾞ 代码,嘻嘻: #include<bits/stdc++.h> using namespace…

计蒜客题目链接:https://nanti.jisuanke.com/t/41303 题目:给你一个序列a,你可以从其中选取元素,构建n个串,每个串的长度为n,构造的si串要满足以下条件, 1. si[1]=i . 2. si[j]<si[j-1] 3. |pos[j] -pos[j-1]|<=k 并且每个a中的元素只能用一次 4. 两个串大小的定义时 前k项相等的前提(k和前面不是一个),Ck>Dk,则C大于D 求Si的长度,并输出 由于Si[ j ] < Si[j - 1] ,…

开始就觉得有思路,结果越敲越麻烦...  题意很简单,就是说一个青蛙从0点跳到m点,最多可以跳l的长度,原有石头n个(都仅表示一个点).但是可能跳不过去,所以你是上帝,可以随便在哪儿添加石头,你的策略是让青蛙跳过去的次数最多,但是你添加了石头后,青蛙会选择最少的次数跳过去,问青蛙跳的次数最多是多少. 原有石头与现在的距离不大于l,就找最远的那个可以跳的石头跳过去.接下来就是主要解决现在的位置与接下来的一块石头的距离大于l的情况了.模拟:上帝开始放石头,要让青蛙一定是这次跳这块石头(在上次不能跳)…

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 finds that Pucci kn…

42.93% 1000ms 131072K LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10^{-9}10−9 second. In the next 10^{-9}10−9 second, he built a data decoding device that decodes data encoded w…

Lattice's basics in digital electronics 44.08% 1000ms 131072K   LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10^{-9}10−9 second. In the next 10^{-9}10−9 second, he built a data d…

Couple Trees Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://hihocoder.com/problemset/problem/1232 Description "Couple Trees" are two trees, a husband tree and a wife tree. They are named because they look like a couple leaning on each other.…

"Couple Trees" are two trees, a husband tree and a wife tree. They are named because they look like a couple leaning on each other. They share a same root, and their branches are intertwined. In China, many lovers go to the couple trees. Under t…

Given a rooted tree ( the root is node 11 ) of NN nodes. Initially, each node has zero point. Then, you need to handle QQ operations. There're two types: 1\ L\ X1 L X: Increase points by XX of all nodes whose depth equals LL ( the depth of the root i…

17.64% 1000ms 131072K   A sequence of integer \lbrace a_n \rbrace{an​} can be expressed as: \displaystyle a_n = \left\{ \begin{array}{lr} 0, & n=0\\ 2, & n=1\\ \frac{3a_{n-1}-a_{n-2}}{2}+n+1, & n>1 \end{array} \right.an​=⎩⎨⎧​0,2,23an−1​−an−…

26.89% 1000ms 131072K 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-emp…

给两颗标号从1...n的树,保证标号小的点一定在上面.每次询问A树上的x点,和B树上的y点同时向上走,最近的相遇点和x,y到这个点的距离. 比赛的时候想用倍增LCA做,但写渣了....后来看到题解是主席树就写了一发 呆马: #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vecto…

题意:给出一个字符串,长度是2*10^4.将它首尾相接形成环,并在环上找一个起始点顺时针或逆时针走一圈,求字典序最大的走法,如果有多个答案则找起始点最小的,若起始点也相同则选择顺时针. 分析:后缀数组. 首先,需要补充后缀数组的基本知识的话,看这个链接. http://www.cnblogs.com/rainydays/archive/2011/05/09/2040993.html 我利用这道题重新整理了后缀数组的模板如下: ; //call init_RMQ(f[], n) first. //…

odd-even number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even nu…

QSC and Master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Every school has some legends, Northeastern University is the same. Enter from the north gate of Northeastern University,You are…

题目链接 Problem Description Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.Given a string s, we define a substring that happens exactly k time…

f(cos(x))=cos(n∗x) holds for all xx. Given two integers nn and mm, you need to calculate the coefficient of x^mx​m​​ in f(x)f(x), modulo 998244353998244353. Input Format Multiple test cases (no more than 100100). Each test case contains one line cons…

题意:给你一棵树,n个点q次操作,操作1查询x子树深度为d的节点权值和,操作2查询子树x权值和 把每个点按(dfn,depth)的二维关系构造kd树,剩下的只需维护lazy标记即可 #include<bits/stdc++.h> #define rep(i,j,k) for(register int i=j;i<=k;i++) #define rrep(i,j,k) for(register int i=j;i>=k;i--) #define erep(i,u) for(regis…

自己太菜,数学基础太差,这场比赛做的很糟糕.本来想吐槽出题人怎么都出很数学的题,现在回过头来想还是因为自己太垃圾,竞赛就是要多了解点东西. 找$f(cos(x))=cos(nx)$中$x^m$的系数模998244353. wolfram alpha查了这个函数无果,得到了一堆sinx和cosx以及一个复指数的方程,其实应该推个几项再用数列查询查查看的,然后就会知道是Chebyshev polynomials 查WIKI直接就有通项公式了.然后就比较简单的了. 连方程都看不出来就别想着推导公式了.…