大意: 给定字符串$s$,$t$, 每次操作可以将$S=AB$变为$S=B^RA$, 要求$3n$次操作内将$s$变为$t$. #include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue>…

Lock Puzzle 题目大意:给你两个字符串一个s,一个t,长度<=2000,要求你进行小于等于6100次的shift操作,将s变成t, shift(x)表示将字符串的最后x个字符翻转后放到最前面. 思路:不会写,看了题解... 因为长度为3000,操作为6500,我们考虑每三次操作将一个字符放到最后,并保证其他字符的顺序不变,这样是可以实现的, 如果我们想要将第k个字符移到最后,我们只要shift(n-1-k) , shift(1) , shift(n-1),就能实现啦 . #includ…

http://acm.hdu.edu.cn/showproblem.php?pid=4708 Rotation Lock Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Alice was felling into a cave. She found a strange door with a number square m…

Rotation Lock Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1836    Accepted Submission(s): 580 Problem Description Alice was felling into a cave. She found a strange door with a numbe…

Rotation Lock Puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 654 Accepted Submission(s): 190 Problem Description Alice was felling into a cave. She found a strange door with a number squar…

Rotation Lock Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 290    Accepted Submission(s): 60 Problem Description Alice was felling into a cave. She found a strange door with a number s…

Rotation Lock Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 695    Accepted Submission(s): 204 Problem Description Alice was felling into a cave. She found a strange door with a number…

Bash and a Tough Math Puzzle CodeForces 914D 线段树+gcd数论 题意 给你一段数,然后小明去猜某一区间内的gcd,这里不一定是准确值,如果在这个区间内改变一个数的值(注意不是真的改变),使得这个区间的gcd是小明所猜的数也算小明猜对.另一种操作就是真的修改某一点的值. 解题思路 这里我们使用线段树,维护区间内的gcd,判断的时候需要判断这个区间的左右子区间的gcd是不是小明猜的数的倍数或者就是小明猜的数,如果是,那么小明猜对了.否则就需要进入这个区间…

题目:http://codeforces.com/contest/936/problem/C 玩了一个小时,只能想出 5*n 的方法. 经过一番观察?考虑这样构造:已经使得 A 串的一个后缀 = B 串的一个前缀,考虑再把一个正确的字符挪到 A 串的最后面. 设该字符为 x .它之前有 len 个字符.当前已经弄好的结尾长度为 l2 . 进行这5个操作:n-len , len , n-len-l2 , l2 , n-l2+1 . 思路就是: 1&2:使得 x 变成结尾: 3:使 x 变成开头.原…

传送门 好久不做构造题脑子都僵化了qwq 无解的条件是\(s\)包含的字符可重集和\(t\)包含的字符可重集不相等,相等的时候下文会给出一种一定可行的构造方案. 考虑增量构造.定义某个字符串\(x\)的反串为\(x'\),设已经构造完成的串为\(S\),\(x\)和\(y\)是即将拼合在\(S\)上的两个字符,\(.\)是其他的无用字符 那么我们通过以下步骤将\(x\)和\(y\)拼合: \(.....x.....S \rightarrow S'..........x \rightarrow x…