Mahmoud was trying to solve the vertex cover problem on trees. The problem statement is: Given an undirected tree consisting of n nodes, find the minimum number of vertices that cover all the edges. Formally, we need to find a set of vertices such th…

解法 很简单对于n<=5举不出反例 如果n>5的话2,3,4好点连1,其他点连2 对于正面例子 直接所有点连1号点 其实就是结论题 代码: #include <cstdio> int main(){ int n; scanf("%d", &n); if (n <= 5) printf("-1\n"); else{ printf("1 2\n"); printf("1 3\n"); prin…

题意:要求构造一个n个数的序列,要求n个数互不相同,且异或结果为x. 分析: 1.因为0 ^ 1 ^ 2 ^ 3 ^ ... ^ (n - 3) ^ (n - 2) ^ (0 ^ 1 ^ 2 ^ 3 ^ ... ^ (n - 3) ^ (n - 2) ^ x) = x, 构造的n个数可以为0,1,2,3,...,(n - 3),(n - 2),(0 ^ 1 ^ 2 ^ 3 ^ ... ^ (n - 3) ^ (n - 2) ^ x). 2.因为(0 ^ 1 ^ 2 ^ 3 ^ ... ^ (n…

题目链接:http://codeforces.com/contest/862/problem/C 题解:一道简单的构造题,一般构造题差不多都考自己脑补,脑洞一开就过了 由于数据x只有1e5,但是要求是1e6,而且我们知道3个数可以组合成任意数也就是说n-3的数从1-1e5直接任意找然后使得其总xor为sum 当sum=x时(定义Max=1<<17 > 1e5)那么这三个数只要组成0就行,当sum!=x时那么这三个数只要构成sum^x就行,这3个数从1e5~1e6中找 很好找的. #inc…

原题链接:http://codeforces.com/contest/862/problem/C 题意:给出n,x,求n个不同的数,使这些数的异或和为x 思路:(官方题解)只有n==2&&x==0时输出NO,接下来考虑YES的情况 先定义一个数pw=217(输出答案时保证不会出现重复数字) 因为 数x 完全可以由三个数异或得到,那么对于n>3的情况,先求1到n-3的异或和,得到y,如果y==x,那么剩下三个数为pw, pw*2, pw^(pw*2)(当x==0时,0, pw,pw^x…

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…

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…

\(\color{#0066ff}{题目描述}\) 给出n个点,n-1条边,求再最多再添加多少边使得二分图的性质成立 \(\color{#0066ff}{输入格式}\) The first line of input contains an integer n - the number of nodes in the tree ( \(1<=n<=10^{5}\) ). The next n−1 lines contain integers u and v ( \(1<=u,v<=…

题目传送门 题目大意:先提供一个数组,让你造一个数组,这个数组的要求是 1 各元素之间都互质  2  字典序大于等于原数组  3 每一个元素都大于2 思路: 1.两个数互质的意思就是没有公因子.所以每确定一个数字之后,就把这个数字的所有公因子全部用vis数组标记一下. 2.每一次找数字都是从a[i]开始找,如果a[i]符合条件则下一个,如果不符合条件就a[i]+1,暴力枚举(贪心),如果有一个地方是大于原数组的,之后的所有数字则从最小的开始找就可以了. 3.找公因子的方法是素数筛法的改编. #i…

C. Mahmoud and Ehab and the xor Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his…