[CF959A]Mahmoud and Ehab and the even-odd game题解

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[CF959A]Mahmoud and Ehab and the even-odd game题解

题意简述一个数n,Mahmoud珂以取(即如果取\(k\),使\(n = n - k\))一个正偶数,Ehab珂以取一个正奇数,一个人如果不能取了(对于Mahmoud和Ehab \(n = 0\),对于Mahmoud \(n = 1\)),对方就赢了. Mahmoud先手,问谁会赢. 解法很简单,如果n是奇数那么Mahmoud一次铁定取不玩,因为奇数减偶数为仍为奇数,所以剩下的肯定是奇数,那么Ehab讲把它取到0必胜:如果n为偶数Mahmoud一次去玩,Ehab必败. 简单来说,就是\(n\…

[CF959F]Mahmoud and Ehab and yet another xor task题解

搞n个线性基,然后每次在上一次的基础上插入读入的数,前缀和线性基,或者说珂持久化线性基. 然后一个num数组记录当时线性基里有多少数然后每次前缀操作一下就珂以了代码 #include <cstdio> #define ll long long const ll MOD = 1e9+7; ll read(){ ll x = 0; int zf = 1; char ch = ' '; while (ch != '-' && (ch < '0' || ch > '9'…

[CF959D]Mahmoud and Ehab and another array construction task题解

解法非常暴力的模拟. 一开始吧\(1 -> 2 \times 10^6\)全部扔进一个set里,如果之前取得数都是与原数组相同的,那么lower_bound一下找到set中大于等于它的数,否则直接取set中最小的数,然后枚举该数的所有质因数及其倍数,类似埃拉托斯特尼筛法删掉它们(如果它们在集合中的话). P.S. 不要用STL algorithm的lower_bound,极慢,用STL内部自带的lower_bound,快很多! 代码 #include <cstdio> #include…

Codeforces 862A Mahmoud and Ehab and the MEX

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…

Codeforces 959F Mahmoud and Ehab and yet another xor task 线性基 (看题解)

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…

E - Mahmoud and Ehab and the bipartiteness CodeForces - 862B (dfs黑白染色)

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…