题目大意: 令conc(a,b)函数得出的结果为将ab拼接得到的数字. 例如:conc(12,23)=1223 a和b不会包括前导0! 接下来,你已知A和B,问有多少对的(a,b)满足 1≤a≤A , 1≤b≤B a*b+a+b=conc(a,b) 解题思路: 想法题,只需要满足b这个数字每一位全为9,那么等式 a*b+a+b=conc(a,b) 恒成立 因为 a*b+a+b=a*(b+1)+b b+1为一个首位是1,其余位全为0的数,且长度为b的长度+1 所以a*(b+1)+b相当于a*(Le…

http://codeforces.com/problemset/problem/442/B (题目链接) 题意 n个人,每个人有p[i]的概率出一道题.问如何选择其中s个人使得这些人正好只出1道题的概率最大. Solution 很显然的概率dp,过了样例即可AC..话说我为什么要刷B题→_→ 代码 // codeforces442B #include<algorithm> #include<iostream> #include<cstdlib> #include<…

Another Rock-Paper-Scissors Problem 题目连接: http://codeforces.com/gym/100015/attachments Description Sonny uses a very peculiar pattern when it comes to playing rock-paper-scissors. He likes to vary his moves so that his opponent can't beat him with hi…

当在Codeforces上做题的时,有时会无意撇到右侧的Problem tags边栏,但是原本并不希望能够看到它. 能否把它屏蔽了呢?答案是显然的,我们只需要加一段很短的CSS即可. span.tag-box{ display: none; } 这样,当这段CSS生效时,标签将会变成如下所示的样子. 这样目的就达到了. 那么,如何将CSS应用到浏览器中使其生效呢?对于不同的浏览器有不同的方法.下面以主流浏览器为例. 对于Safari,将上述代码另存为.css文件,然后在偏好设置->高级->样式…

这道题是这种,给主人公一堆事件的成功概率,他仅仅想恰好成功一件. 于是,问题来了,他要选择哪些事件去做,才干使他的想法实现的概率最大. 我的第一个想法是枚举,枚举的话我想到用dfs,但是认为太麻烦. 于是想是不是有什么规律,于是推导了一下,推了一个出来,写成代码提交之后发现是错的. 最后就没办法了,剩下的时间不够写dfs,于是就放弃了. 今天看thnkndblv的代码,代码非常短,于是就想肯定是有什么数学规律,于是看了一下, 果然如此. 是这种,还是枚举,当然是有技巧的,看我娓娓道来. 枚举的话…

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…

题目链接:http://codeforces.com/problemset/problem/803/G 大致就是线段树动态开节点. 然后考虑到如果一个点还没有出现过,那么这个点显然未被修改,就将这个点所代表的区间定位到原序列中,利用ST表查一下区间最小值就可以了. 定位: llg minn(llg l,llg r) { >=n) return mt; l%=n;if(!l) l=n; r%=n;if(!r) r=n; ,r)); else return gw(l,r); } 其中${gw(l,r…

链接 [http://codeforces.com/contest/1076/problem/C] 题意 a+b=d and a⋅b=d. 计算出a和b 分析 ab=a(d-a)=d aa-ad+d=0; 先判断再直接求根公式 代码 #include<bits/stdc++.h> using namespace std; int main(){ int t,d; cin>>t; while(t--){ cin>>d; if(d*d-4*d<0) cout<&…

题目传送门 传送门I 传送门II 传送门III 题目大意 给定一个网络.网络分为$A$,$B$两个部分,每边各有$n$个点.对于$A_{i} \ (1\leqslant i < n)$会向$A_{i + 1}$连一条容量为$x_{i}$的有向边,对于$B_{i} \ (1\leqslant i < n)$会向$B_{i + 1}$连一条容量为$y_{i}$的有向边.$A$和$B$之间有$m$条边,起点为$A_{u_{i}}$,终点为$B_{v_{i}}$,容量为$w_{i}$的有向边.要求支持…

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…

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…

C. Mike and gcd problem time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful…

题目链接:http://codeforces.com/problemset/problem/798/C 题意:给你n个数,a1,a2,....an.要使得gcd(a1,a2,....an)>1,可以执行一次操作使ai,ai+1变为ai - ai + 1, ai + ai + 1.求出使得gcd(a1,a2,....an)>1所需要的最小操作数. 解题思路:首先,要知道如果能够实现gcd(a1,a2,....an)>1,那么a1~an肯定都是偶数(0也是偶数),所以我们的目的就是用最少的操…

[题目链接]:http://codeforces.com/contest/776/problem/D [题意] 每个门严格由两个开关控制; 按一下开关,这个开关所控制的门都会改变状态; 问你能不能使所有的门都打开(同时); [题解] /* 对于每个门, 有两个开关连接着它; 用一条边连接这两个开关 则如果这个门的状态是关的, 则这条边的边权为1 表示它们俩的颜色不能一样; (也即这两个开关只能改变一个) 也即要1开1关; 如果这个门的状态是开的 则这条边的边权为0: 表示它们俩的颜色相同; 即同…

http://codeforces.com/problemset/problem/776/D 题意:有n个门,m个开关,每个门有一个当前的状态(0表示关闭,1表示打开),每个开关控制k个门,但是每个门确切的受两个开关控制,如果一个开关打开,那么原来关闭的门会打开,打开的门关闭,问是否存在一个情况使得所有的门打开. 思路:类似于01染色,把开关当成点,门当前的状态当成边权建图.初始先假设一个门的状态(初始假设为0和假设为1都是一样的,举几个例子就发现了),然后因为每个门受两个开关控制,所以可以推出…

Time limit 2000 ms Memory limit 262144 kB Source Educational Codeforces Round 69 (Rated for Div. 2) Tags dp greedy math *1900 Editorial Announcement (en) Tutorial #1 (en) Tutorial #2 (en) Tutorial #3 (ru) 官方题解 At first let's solve this problem when m…

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…

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…