AT2292 Division into Two】的更多相关文章

题目 不妨认为\(A>B\). 首先判一下无解. 设\(f_i\)表示\(A\)集合最后选第\(i\)个数的方案数. 转移的话枚举一下从哪个\(j\)转移过来. 显然\(j\)需要满足以下条件: \(j<i\) \(S_j<S_i-A\) \(\forall a,b\in(j,i)\wedge a<b,S_a<S_b-B\) 不难发现\(j\)的取值范围是一个区间,前缀和优化即可. #include<cstdio> #include<cctype> #…
正解:$dp$ 解题报告: 传送门$QwQ$ 不妨令$A\geq B$,于是先$sort$然后预处理判下如果有三个元素两两差都小于$B$的就直接$GG$了. 然后考虑对集合$X$进行$dp$,剩下的数放到$Y$就成.设$f_i$表示集合$X$最后一个选择的是$i$时的方案数,就有$f_i=\sum f_j,a_i-a_j\geq A$. 嗷然后因为我很呆所以我开始想的时候漏了个条件把代码打完才发现样例都过不去,,,就还需要$[j+1,i-1]$中任意两个数的差大于等于$B$.显然$j$就一段区间…
传送门 Luogu 解题思路 考虑如何 \(\text{DP}\) 为了方便处理,我们设 \(A > B\) 设 \(dp[i]\) 表示处理完 \(1...i\) ,并且第 \(i\) 个数放入关于 \(A\) 的集合中的方案. 转移就只需要枚举前一个数 \(j\) 就好了. 但是观察到 \(N \le 10^5\) ,我们就需要考虑优化. 观察到每次 \(j\) 的取值都是一段连续的区间,所以我们可以前缀和优化一下,就可以做到 \(O(n)\) 细节注意事项 咕咕咕 参考代码 #includ…
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 =…
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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…
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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 =…