Codeforces Round #313 D. Equivalent Strings（DFS）

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Codeforces Round #313 D. Equivalent Strings（DFS）

D. Equivalent Strings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal l…

Codeforces Round 662 赛后解题报告（A-E2）

Codeforces Round 662 赛后解题报告梦幻开局到1400+的悲惨故事 A. Rainbow Dash, Fluttershy and Chess Coloring 这个题很简单,我们可以画几张图,发现每一次我们染色的最佳方法就是每次往里面多填一圈,并把上一圈给填满. 比如上图就很好地说明了这个过程,大家可以用画一下 \(n=4,n=5,n=6,n=7\),就能验证这个命题了,所以一个 \(n\times n\) 的矩阵有 \(\lfloor\frac{n}{2}\rfloor+…

Codeforces Round #306 (Div. 2) ABCDE（构造）

A. Two Substrings 题意:给一个字符串,求是否含有不重叠的子串"AB"和"BA",长度1e5. 题解:看起来很简单,但是一直错,各种考虑不周全,最后只能很蠢的暴力,把所有的AB和BA的位置求出来,能有一对AB和BA不重叠即可. #include <bits/stdc++.h> using namespace std; ]; vector<int> ab; vector<int> ba; int main() { w…

CodeForces 165C Another Problem on Strings（组合）

A string is binary, if it consists only of characters "0" and "1". String v is a substring of string w if it has a non-zero length and can be read starting from some position in string w. For example, string "010" has six sub…

Codeforces Round #527 (Div. 3)F（DFS,DP）

#include<bits/stdc++.h>using namespace std;const int N=200005;int n,A[N];long long Mx,tot,S[N];vector<int>Adj[N];void DFS(int v,int p){ S[v]=A[v]; for(int &u:Adj[v]) if(u!=p) DFS(u,v),S[v]+=S[u]; if(p)//0号结点是不存在的…

Codeforces Round #267 (Div. 2)D（DFS+单词hash+简单DP）

D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the E…