C++,codeforces,1873A, A. Short Sort /* codeforces,1873A, A. Short Sort 判断一个由a,b,c组成的字符串是否是可以通过对字符串"abc"进行1次或0次交换字符操作得到 交换字符操作是指,将字符串中的两个字符交换位置 */ /* 符号条件的字符串一定有一个位置的字符与"abc"相同,即 abc,acb abc,cba abc,...
A. Short Sort 【交换】 描述:对一个字符串“acb”/“bac”……选择任意两个字符做交换,若最后能得到“abc”就输出YES。 解: 特解:因为只交换两个,剩下那一个就必须在原位置。 strings;cin>>s;if(s[0]=='a '||s[1]=='b '||s[2]=='c')YY;elseNN; 通解:如果要模拟交换,就会有Cn2=n(n...
Short Sort * Contest: Codeforces - Codeforces Round 898 (Div. 4) * URL: https://codeforces.com/contest/1873/problem/A * MemoryL: 256 MB * TimeL: 1000 ms * === */ #pragma GCC optimize("Ofast") #include<bits/stdc++.h> #define IOS ios::sync_with_stdio(false),cin.tie(nullptr)...
Div2-Medium:http://ideone.com/cazTBB Div2-Hard:http://ideone.com/rxEUcR Div1-Easy:http://ideone.com/JLuDVM Div1-Medium:http://ideone.com/7FEz1K Div1-Hard:http://ideone.com/clL4Rk Editorial: Div1-Easy: The key is that: Any tree is a bipartite graph! That means, if two nodes...
Typical applications of the suffix automaton are based on some sort of dynamic programming scheme on top of the automaton, but let's start with the trivial ones that can be implemented on the plain automaton. Problem:Find whether a given stringwis a substring ofs. ...
We know that if there's a solution, then the heights from the min. to max. height are non-decreasing on each side. If there's another solution, we can sort the foxes on each side between the min. and max. height without worsening the max. difference. ...
Sort all skill levels, and for each value find the number of students with such skill level. Note that if the team contains students with skill levels l and r, all students with skill levels m such that l < m < r will also be in the same team. So let us calculate the ...