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…

Codeforces Round #297 (Div. 2)D. Arthur and Walls Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx  Solved: 2xx 题目连接 http://codeforces.com/contest/525/problem/D Description Finally it is a day when Arthur has enough money for buying an apartment. H…

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…

Problem - C - Codeforces 看清题目要求, 最重要部分在第二段. 大佬最后给出的代码果然简单, 思路简单化, 未必非要把答案在一个大括号里全部完成, 两个指针同时跑,中间加了一堆判断处理, 反而不如返璞归真, 简单清晰为后面改代码也省了很多力气 AC代码 #include <iostream> #include <algorithm> using namespace std; typedef long long LL; const int N = 2e5+10…

Problem - D1 - Codeforces 题意: 给n个符号(+或-), +代表+1, -代表-1, 求最少删去几个点, 使得     题解(仅此个人理解): 1. 这题打眼一看, 肯定和奇偶有关系, 奇数为+, 偶数为-, 但是删去点这一操作是动态的, 删去某点后, 后面的点的正负随之颠倒, 即奇数位+变偶数位-, 偶数位-变奇数位+, 恰好可以利用该特质 也就是说可以这样理解: 找到一个位置,删去该点并使得后面的点正负颠倒, 最后满足条件. 2. 对于奇数个数, 必须变成偶数个个数…

题目链接  Round 439 div2 就做了两道题TAT 开场看C题就不会 然后想了好久才想到. 三种颜色挑出两种算方案数其实是独立的,于是就可以乘起来了. E题想了一会有了思路,然后YY出了一种方案. 我们可以对每个矩形随机一个权值,然后用二维树状数组搞下. 询问的时候看两个点权值是否相等就可以了 于是就过了. D题待补 给出一棵完全二叉树,这棵树上有附带的m条边(m <= 4),求这张图的简单路径条数. qls的题就是厉害…… C题 #include <bits/stdc++.h>…

http://codeforces.com/contest/426/problem/B 对称标题的意思大概是.应当指出的,当线数为奇数时,答案是线路本身的数 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 1005 using namespace std; char a[maxn][maxn]; int ans; void solve…

Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as f…

The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c). Over time the stars twinkle. At moment 0 the i-th…

Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of s characters. The first participant types one character in v1 milliseconds and has ping t1 milliseconds. The secon…

Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other. The boys decided to h…

n people are standing on a coordinate axis in points with positive integer coordinates strictly less than 106. For each person we know in which direction (left or right) he is facing, and his maximum speed. You can put a bomb in some point with non-n…

It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy to Petya, but he thinks he lacks time to finish them all, so he asks you to help with one.. There is a glob pattern in the statements (a string consis…

It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends. Today he invented one simple game to play with Lena, with whom he shares a desk. The rules…

A. The Artful Expedient 题目链接:http://codeforces.com/contest/869/problem/A 题目意思:给你两个数列,各包含n个数,现在让你从上下两个数列中各取一个数a[i],b[j],如果a[i]^b[j]在这2×n个数里面出现过,那么就获得一分,问将任意的a[i],b[j]之间的亦或之后,如果分数是奇数则Koyomi胜利,否则Karen胜利.问最后到底谁胜了. 题目思路:非常无聊的题目,暴力都可以过,就是暴力枚举a[i],b[j],把所有答…

E. The Untended Antiquity 题目链接http://codeforces.com/contest/869/problem/E 解题心得: 1.1,x1,y1,x2,y2 以(x1,y1)为左上角(x2,y2)为右下角的矩形,四边建墙.      2,x1,y1,x2,y2 以(x1,y1)为左上角(x2,y2)为右下角的矩形,拆除墙(肯定存在).      3,x1,y1,x2,y2问是否可以从(x1,y1)走到(x2,y2)(没墙阻隔). 2.先要标记每个点的状态,将用墙…

C. The Intriguing Obsession 题目链接http://codeforces.com/contest/869/problem/C 解题心得:     1.由于题目中限制了两个相同颜色的点之间长度至少为3,这样就只能两两不同颜色的点相互组合,再将三种组合情况的可能种数乘起来.     2.然后就是两个不同颜色的点怎么组合的问题,这就是一个记忆化搜索(DP),假如红色点和黄色点相互组合,这样一个红色点可以对应一个黄色点(dp[x][y] = dp[x-1][y-1]*y),或者…

B. The Eternal Immortality 题目链接http://codeforces.com/contest/869/problem/B 解题心得:题意就是给出a,b,问(a!)/(b!)的个位数,要注意0,5两个数,只要a,b相差超过5个位数就只能是0,其实没有看到相差5看到相差10也可以的,然后又暴力跑一个末位数就可以了. /*这里跑的是相差10位*/ #include<bits/stdc++.h> using namespace std; typedef long long…

A. The Artful Expedient 题目链接http://codeforces.com/contest/869/problem/A 解题心得:就是一个水题,读懂题就好,题意是,(i,j)ai异或bi,统计ai异或bi得到的数能够在ai或bi中找的个数,如果是偶数输出Karen,否则输出Karen. #include<bits/stdc++.h> using namespace std; const int maxn = 20000; const int maxn2 = 2e7+10…

http://codeforces.com/problemset/problem/590/A: 在CF时没做出来,当时直接模拟,然后就超时喽. 题意是给你一个0 1串然后首位和末位固定不变,从第二项开始到倒数第二项,当前的a[i]=(a[i-1],a[i],a[i+1])三项排序后的中间项,比如连续3项为 1 0 1,那么中间的就变为1,然后题目让你输出达到稳定状态时所需的最小步数,不能的话输出-1. 无论给你啥数列,都能达到稳态.所以不可能输出-1: 还有一开始就稳定不变,或经过几次变换而稳定…

2018年11月30日更新,补充了一些思考. 题意(CodeForces 869C) 三堆点,每堆一种颜色:连接的要求是同色不能相邻或距离必须至少3.问对整个图有几种连接方法,对一个数取模. 解析 要求很重要:同色不能相邻很容易理解,但是>=3比较难理解.比较常见的是R->G->B->R,这样能看出来一个重要的结论:对单个节点只能连接某个颜色至多一个点(不然一定有距离为2的点).这样一来我们思考一下状态会和哪些东西有关联:如果我放一个一个新颜色在里面,它会怎么同原图产生联系?一是和…

A. DZY Loves Hash time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert n numbers, in the order they are given, i…