题意: 给定一系列字符串,每次都是后一个字符串和前面的融合,这个融合操作就是原来的串分成独立的,然后把新串插入到这些空格中.问最后,最长的相同连续的长度. 思路: 这道题可以贪心的来,我们压缩状态,记录串中每个字母对应最长的长度.然后分类讨论处理就行了. #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib>…

E. String Multiplication 题意 分析: 从后往前考虑字符串变成什么样子. 设$S_i = p_1 \cdot p_2 \dots p_{i}$,最后一定是$S_{n - 1} \cdot p_n$,就是将$S_{n-1}$每两个字符之间放入$p_n$.按照$p_n$分类讨论,那么最后有三种情况. 设$p_n$的开头字符是$c0$,结尾字符是$c1$,包含开头的连续段的长度是$len0$,包含结尾的连续段的长度是$len1$. 1.$c0 \neq c1$,那么答案可以是三…

Problem   Codeforces #541 (Div2) - E. String Multiplication Time Limit: 2000 mSec Problem Description Input Output Print exactly one integer — the beauty of the product of the strings. Sample Input 3aba Sample Output 3 题解:这个题的思维难度其实不大,需要维护什么东西很容易想到,或…

原题地址: 344 Reverse String: https://leetcode.com/problems/reverse-string/description/ 541 Reverse String II: https://leetcode.com/problems/reverse-string-ii/description/ 题目&&解法: 1.Reverse String: Write a function that takes a string as input and ret…

344. Reverse String 最基础的旋转字符串 class Solution { public: void reverseString(vector<char>& s) { if(s.empty()) return; ; ; while(start < end){ char tmp = s[end]; s[end] = s[start]; s[start] = tmp; start++; end--; } return; } }; 541. Reverse Strin…

package leadcode; /** * 541. Reverse String II * Easy * 199 * 575 * * * Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters lef…

E. String Multiplication time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and h…

problem 541. Reverse String II 题意: 给定一个字符串,每隔k个字符翻转这k个字符,剩余的小于k个则全部翻转,否则还是只翻转剩余的前k个字符. solution1: class Solution { public: string reverseStr(string s, int k) { int n = s.size(); int cnt = n / k; ; i<=cnt; i++) { ==) { if(i*k+k<n) reverse(s.begin()+i…

这题难度2200,应该值了. 题目链接:CF原网 题目大意:定义两个字符串 $s$ 和 $t$($s$ 的长度为 $m$)的乘积为 $t+s_1+t+s_2+\dots+t+s_m+t$.定义一个字符串的美丽度为最长的相同字母连续子序列的长度.现在给出 $n$ 个字符串 $p_i$,问 $((p_1p_2)p_3)\dots p_n$ 的美丽度. $1\le n\le 10^5,\sum|p_i|\le 10^5$. 官方题解讲的很复杂,但看起来也就是个暴力大模拟,跟我的做法差不多. 为叙述方便…

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or eq…