A. Gudako and Ritsuka 链接 by Yuki & Asm.Def 期望难度:Hard- 考虑从后往前进行博弈动态规划,在这一过程中维护所有的先手必胜区间.区间不妨采用左开右闭,方便转移. 考虑一次转移,如果当前Servant的后一个位置属于对手,则当前Servant的必胜区间可以通过将后一个Servant的每个必败区间的左端点+1.右端点+x得到:如果后一个位置属于自己,则可以通过将后一个Servant的必胜区间做同样的操作得到.不妨分别对必胜区间左右端点维护一个偏移量,需要…
https://nanti.jisuanke.com/t/31452 题意 给出一个n (2 ≤ N ≤ 10100 ),找到最接近且小于n的一个数,这个数需要满足每位上的数字构成的集合的每个非空子集组成的数字是个素数或1. 分析 打表发现满足要求的数字很少.实际上因为一个数不能出现两次,而偶数不能存在.这样最后只有20个数符合要求. #include <bits/stdc++.h> using namespace std; typedef long long ll; ; ; ] = {,,,…
Supreme Number 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 N as the supreme number if and only if each number made up of an non-empty subse…
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 N spots in the jail and MM roads connecting some of the spots. JOJO finds…
131072K   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 tha…
J. Ka Chang 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 o…
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…
"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…
题意: 二分图 有k条边,我们去选择其中的几条 每选中一条那么此条边的u 和 v的度数就+1,最后使得所有点的度数都在[l, r]这个区间内 , 这就相当于 边流入1,流出1,最后使流量平衡 解析: 这是一个无源汇有上下界可行流 先添加源点和汇点 超级源超级汇  跑遍dinic板子 就好了...看了一发蔡队的代码,学到了好多新知识(逃)... #include <bits/stdc++.h> #define mem(a, b) memset(a, b, sizeof(a)) #define r…
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…