题意:求定 n 个数,求有多少对数满足,ai^bi = x. 析:暴力枚举就行,n的复杂度. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <c…

给出N*M矩阵,对于该矩阵有两种操作: 1.交换两列,对于整个矩阵只能操作一次 2.每行交换两个数. 交换后是否可以使每行都递增. *解法:N与M均为20,直接枚举所有可能的交换结果,进行判断 每次枚举注意列和行都可以选择不交换 #include <iostream> #include <cstdio> using namespace std; int n, m; ][]; bool check(int t)//检查第t行是否符合递增 { ; ; i < m; i++) ])…

题目链接:http://codeforces.com/contest/742/problem/B B. Arpa's obvious problem and Mehrdad's terrible solution time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are some beautiful girls in…

B. Arpa’s obvious problem and Mehrdad’s terrible solution time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are some beautiful girls in Arpa’s land as mentioned before. Once Arpa came u…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output There are some beautiful girls in Arpa's land as mentioned before. Once Arpa came up with an obvious problem: Given an array and a number x, coun…

There are some beautiful girls in Arpa’s land as mentioned before. Once Arpa came up with an obvious problem: Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that , where is bitwise xor operation (see not…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output   There are some beautiful girls in Arpa’s land as mentioned before. Once Arpa came up with an obvious problem: Given an array and a number x…

Content 有一个长度为 \(n\) 的数组,请求出使得 \(a_i \oplus a_j=x\) 且 \(i\neq j\) 的数对 \((i,j)\) 的个数.其中 \(\oplus\) 表示异或符号. 数据范围:\(1\leqslant n,a_i\leqslant 10^5,0\leqslant x\leqslant 10^5\). Solution 利用一个异或的性质:\(a\oplus b=c\Rightarrow a\oplus c=b,b\oplus c=a\),我们发现问题…

codeforces 741D Arpa's letter-marked tree and Mehrdad's Dokhtar-kosh paths 题意 给出一棵树,每条边上有一个字符,字符集大小只有22. 对于每一个子树,询问其中最长的,满足:路径上的字符集可以重组成字符串的路径的长度. 题解 明显是用mask维护信息,然后启发式合并一下. 一般启发式合并需要用map维护信息,这样的复杂度是log^2的.如果保留每个点重儿子的信息,就可以用全局变量维护,全局变量的大小就可以开很大,可以做到l…

题目链接:Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths 第一次写\(dsu\ on\ tree\),来记录一下 \(dsu\ on\ tree\)主要维护子树信息,往往可以省掉一个数据结构的启发式合并.大体思路如下: 轻重链路径剖分之后,对每个点先递归处理他的所有轻儿子,每次处理完轻儿子之后把这棵子树的信息清空.最后再来处理重孩子,重儿子的信息就可以不用清空了.由于我们是用一个全局数组来记录信息的,重儿子子树的信息就仍然保留…