[AGC009C]Division into 2】的更多相关文章

题意: 有一个长度为$N$的递增序列$S_i$,要把它分成$X,Y$两组,使得$X$中元素两两之差不小于$A$且$Y$中元素两两之差不小于$B$,求方案数 首先考虑$O\left(n^2\right)$的做法: 为了方便,我们令$S_0=-\infty$ 设$f_{M,i,j}(M\in\{X,Y\},1\leq i\leq n,0\leq j\lt i)$表示已划分好$S_{1\cdots i}$且$S_j$是最后一个不属于$M$的元素的方案数 已算好$f_{X,1\cdots i,j}$和$…
题意 有\(n\)个严格升序的数,请你分成两个集合\(A\)和\(B\),其中一个集合任意两数之差不小于\(x\),另一集合任意两数之差不小于\(y\). 问方案数,集合可以为空. $n \le 10^5 $ 传送门 思路 又是一道神仙\(dp\) 设\(dp_i\)表示当前\(B\)集合的最后一个数是\(a_i\)的方案数. 如果暴力转移就是:\[dp_i=\sum_{j<i \& a_i-a_j\ge y}dp_j\] 并且满足区间\([j+1,i-1]\)能够放在\(A\)集合中 可以…
[AGC009C]Division into Two 题面 洛谷 题解 首先有一个比较显然的\(n^2\)算法: 设\(f_{i,j}\)表示\(A\)序列当前在第\(i\)个,\(B\)序列当前在第\(j\)个的方案数,发现\(i,j\)大小没有限制不是很好转移,于是再设一个\(g_{i,j}\)表示\(B\)序列当前在第\(i\)个,\(A\)序列当前在第\(j\)个的方案数的,这样子我们就可以钦定\(i>j\)了,转移会方便许多. 转移如下: \[ \left\{ \begin{align…
1.在python2 中导入未来的支持的语言特征中division(精确除法),即from __future__ import division ,当我们在程序中没有导入该特征时,"/"操作符执行的只能是整除,也就是取整数,只有当我们导入division(精确算法)以后,"/"执行的才是精确算法. 如: #python 2.7.6 Python 2.7.6 (default, Nov 10 2013, 19:24:18) [MSC v.1500 32 bit (In…
Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0. Example:Given a / b =…
Did you ever have the feeling that adding people doesn't help in software development? Did you ever think about the reason? And do you have any idea to make a change? Traditional software engineering emphasizes on division of work by modules, values…
Contestants Division   Description In the new ACM-ICPC Regional Contest, a special monitoring and submitting system will be set up, and students will be able to compete at their own universities. However there’s one problem. Due to the high cost of t…
题目传送门 /* 暴力:对于每一个数都判断,是否数字全都使用过一遍 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <set> #include <queue> usin…
GDC2016[全境封锁(Tom Clancy's The Division)]对为何对应Eye Tracked System,以及各种优点的演讲报告 原文 4Gamer編集部:松本隆一 http://www.4gamer.net/games/241/G024173/20160317137/         北美时间2016年3月16日,在旧金山举办的Games Developers Conference 2016上,进行了关于[全境封锁(Tom Clancy's The Division)]的…
Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0. Example: Given a / b =…