题意:给你两个串,S2是S1 的一个子串(可以不连续).给你一个s1字符下标的一个排列,按照这个数列删数,问你最多删到第几个时S2仍是S1 的一个子串. 题解:二分删掉的数.判定函数很好写和单调性也可以维持. 不知道为啥答案要-1..貌似l最终指向删掉以后不再保持性质的第一个数. #define _CRT_SECURE_NO_WARNINGS #include<cstdio> #include<algorithm> #include<iostream> #include…

题目链接:http://codeforces.com/problemset/problem/779/D 题意:有两个字符串一个初始串一个目标串,有t次机会删除初始串的字符问最多操作几次后刚好凑不成目标串 所得结果减1 其实简单匹配只需要n只要二分查找一下答案那么结果就是n*logn #include <iostream> #include <cstring> #include <string> using namespace std; const int M = 2e5…

Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her. Sergey gives Nastya the word t and wa…

time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, b…

D. String Game time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty…

D. String Game time limit per test:2 seconds memory limit per test:512 megabytes input:standard input output:standard output Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty…

Link 题意: 给出两字符串$a$,$b$及一个序列,要求从前往后按照序列删掉$a$上的字符,问最少删多少使$b$串不为a的子串 思路: 限制低,直接二分答案,即二分序列位置,不断check即可. /** @Date : 2017-05-07 20:26:33 * @FileName: 779D 二分答案.cpp * @Platform: Windows * @Author : Lweleth (SoundEarlf@gmail.com) * @Link : https://github.co…

题目链接:http://codeforces.com/contest/779/problem/D 题意:给你一段操作序列,按顺序依次删掉字符串1中相应位置的字符,问你最多能按顺序删掉多少个字符,使得s2是剩下的字符构成的字符串的子列. 字符串长度为2e5每次查询的时间复杂度为n如果直接暴力那么复杂度就是n*n 如果二分一下答案的话复杂度就是n*logn再加上修改的复杂度总的复杂度是 (n+n)* logn #include <iostream> #include <cstring>…

[题目链接]:http://codeforces.com/contest/779/problem/D [题意] 给你一段操作序列; 按顺序依次删掉字符串1中相应位置的字符; 问你最多能按顺序删掉多少个字符; 使得s2是剩下的字符构成的字符串的子列; [题解] 二分枚举能够按顺序删掉多少个字符m; 然后把1..m相应的字符标记成已经删掉了; 然后O(N)判断s2是不是剩下的字符的子串; 心态炸了. [完整代码] #include <bits/stdc++.h> using namespace s…

http://codeforces.com/problemset/problem/778/A 题意:给出字符串s和字符串p,还有n个位置,每一个位置代表删除s串中的第i个字符,问最多可以删除多少个字符使得s串依旧包含p串. 思路:想到二分,以为二分做法依旧很暴力.但是别人的做法确实就是二分暴力搞啊. 枚举删除字符数,然后判断的时候如果s串包含p串,那么可以往右区间找,否则左区间找. #include <bits/stdc++.h> using namespace std; #define N…