Adieu l'ami. Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around the abandoned Eikou Cram School building, Oshino's makeshift residence. The space is represented by a rectangular grid of n × m cells, arranged into n…

— This is not playing but duty as allies of justice, Nii-chan! — Not allies but justice itself, Onii-chan! With hands joined, go everywhere at a speed faster than our thoughts! This time, the Fire Sisters — Karen and Tsukihi — is heading for somewher…

Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lif…

Rock... Paper! After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows. A positive integer n is decided first. Both Koyomi a…

Codeforces Round #439 (Div. 2) codeforces 869 A. The Artful Expedient 看不透( #include<cstdio> int main(){ puts("Karen"); ; } 15ms codeforces 869B. The Eternal Immortality(数学,水) 题意:输出两个数的阶乘的商的 个位数 题解:两数之差大于5,个位数就是0.小于5直接个位相乘即可. #include<cs…

& -- 位运算之一,有0则0 原题链接 Problem - 1514B - Codeforces 题目 Example input 2 2 2 100000 20 output 4 226732710 题意 t个测试样例,在每个样例中 数组有n个数,数字范围[ 0, 2k - 1] 使得数组每个数&后,结果=0,并且这n个数的和要尽量大 输出有多少个这样的数组 解析 数组每个数&后,结果=0  -->每一位至少一个0 数要尽量大  -->只要这一位可以 != 0, 就…

还是看大佬的题解吧 CFRound#753(Div.3)A-E(后面的今天明天之内补) - 知乎 (zhihu.com) 传送门  Problem - D - Codeforces 题意 n个数字,n个字符, 数字与字符一一对应, 其中字符R红色代表数字可以上升, B蓝色代表可以下降, 问一顿操作下来能否使n个数含有1~n的所有数 题解 当时真的没想到, 蓝色一定在数组前面, 红色一定在数组后面,( 反正就算有方案使得蓝不在前面, 红不在后面,蓝红也是可以交换的). 排序后, 前面的蓝色数字如果…

Problem - B - Codeforces 就是给你个序列, 给他整成升序的, 每次操作可以使相邻两个数交换位置, 交换条件是二数之和为奇数 结果只需输出是否可以整成升序的 思路: 需要奇数偶数分开讨论, 如果奇数和偶数都分别是单增的那么可行, 反之为no #include <bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<int,int> PII; const int N = 1e…

题意: 给出n个数字,要求在这n个数中选出至少两个数字,使得它们的和在l,r之间,并且最大的与最小的差值要不小于x.n<=15 Problem - 550B - Codeforces 二进制 利用二进制, 第i位为1则加上a[i], 为0则不加, #include<iostream> #include <algorithm> #include <cmath> #include<map> using namespace std; typedef long…

Problem - A - Codeforces 题目 题意很简单每次操作可以使得a1 a2  a3任意两个数分别+1  -1 求最后使得a+c-2b绝对值的最小值 BUG就是最后忽略了-2和2这一点, 他们再进行一次操作就可以变成1和-1, 最后的绝对值也就是1, 所以最后的答案没有2  ! ! ! 官方代码 #include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(0); cin…