Description 给你一个字符串ss,共有qq次操作,每个都是下面两种形式的一种. 11 ii cc 这个操作表示将字符串ss的第ii项变为字符cc 22 ll rr yy 这个操作表示输出字符串yy在字符串ss中以第ll项为起点,以第rr项为终点的子串(包括第ll和第rr项)中作为子串出现的次数. Solution 和BZOJ 4503一样 稍微改改 Code #include <bitset> #include <string> #include <stdio.h&…

题意 题目链接 Sol Orz jry 和上一个题一个思路吧,直接bitset乱搞,不同的是这次有了修改操作 因为每次修改只会改两个位置,直接暴力改就好了 #include<bits/stdc++.h> using namespace std; const int MAXN = 1e5 + 10; char s[MAXN], tmp[MAXN]; int N, q; bitset<MAXN> ans, B[27]; main() { scanf("%s %d",…

[题目]F. Substrings in a String [题意]给定小写字母字符串s,支持两种操作:1.修改某个位置的字符,2.给定字符串y,查询区间[l,r]内出现y多少次.|s|,Σ|y|<=10^5,time=6s. [算法]Bitset [题解]假设S的长度为n,那么对每个字符建一个长度为n的bitset,1表示该位置为该字符,修改时直接修改. 查询的时候将字符串y所有的字符的bitset,按顺序错位and,这样最后得到1表示y为子串,count一下即可. 复杂度O(n^2/32),…

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". Now we have another string p. Your job is to find…

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". Now we have another string p. Your job is to find…

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". Now we have another string p. Your job is to find…

2018-09-01 22:50:59 问题描述: 问题求解: 如果单纯的遍历判断,那么如何去重保证unique是一个很困难的事情,事实上最初我就困在了这个点上. 后来发现是一个动态规划的问题,可以将每个字符结尾的最长长度进行保存,这样就巧妙的解决的重复的问题. The max number of unique substring ends with a letter equals to the length of max contiguous substring ends with that…

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". Now we have another string p. Your job is to find…

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "-zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd-.". Now we have another string p. Your job is to find out…

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". Now we have another string p. Your job is to find…