A. Mahmoud and Ehab and the MEX time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in In…

传送门:CF-862A A. Mahmoud and Ehab and the MEX time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Ol…

第一反应是打表找规律……(写了个prim)但是太菜了没找到 于是开始怀疑是不是我的表错了,又写了一个克鲁斯卡尔,然后结果是一样的……(捂脸) 后来从克鲁斯卡尔的算法上发现了一点东西,发现只有2的幂次长度的边才会对答案产生贡献.假设有n个点,那么就相当于把n - 1的点连入MST,打表找规律可得一个二进位位权值的边会把当前一半的点连入MST,于是水过 Code: #include <cstdio> using namespace std; typedef long long ll; ll n,…

[链接]h在这里写链接 [题意] 在这里写题意 [题解] 让x没有出现,以及0..x-1都出现就可以了. [错的次数] 0 [反思] 在这了写反思 [代码] #include <bits/stdc++.h> using namespace std; const int N = 100; int a[N + 10], n, x; int main() { //freopen("F:\\rush.txt", "r", stdin); ios::sync_wi…

You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty. For exampl…

E. Mahmoud and Ehab and the xor-MST https://codeforces.com/contest/959/problem/E 分析: 每个点x应该和x ^ lowbit(x)连边,那么现在就是求$\sum_{i=1}^{n}lowbit(i)$,然后打表找规律. 代码: #include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #in…

Mahmoud and Ehab and yet another xor task 存在的元素的方案数都是一样的, 啊, 我好菜啊. 离线之后用线性基取check存不存在,然后计算答案. #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair…

Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees. A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in su…

862B - Mahmoud and Ehab and the bipartiteness 思路:先染色,然后找一种颜色dfs遍历每一个点求答案. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) ; bool color[N]; vector<int>g[N]; ; int…

862C - Mahmoud and Ehab and the xor 思路:找两对异或后等于(1<<17-1)的数(相当于加起来等于1<<17-1),两个再异或一下就变成0了,0异或x等于x.所以只要把剩下的异或起来变成x就可以了.如果剩下来有3个,那么,这3个数可以是x^i^j,i,j. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back…