tricky c++ new(this)

Codefoces 429 D. Tricky Function

裸的近期点对.... D. Tricky Function time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an imp…

Codeforces Round #245 (Div. 1) 429D - Tricky Function 最近点对

D. Tricky Function Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 codeforces.com/problemset/problem/429/D Description Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The sele…

Educational Codeforces Round 1 A. Tricky Sum 暴力

A. Tricky Sum Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/598/problem/A Description In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.…

算法训练 Tricky and Clever Password

算法训练 Tricky and Clever Password 时间限制:2.0s 内存限制:256.0MB 问题描述在年轻的时候,我们故事中的英雄——国王 Copa——他的私人数据并不是完全安全地隐蔽.对他来说是,这不可接受的.因此,他发明了一种密码,好记又难以破解.后来,他才知道这种密码是一个长度为奇数的回文串. Copa 害怕忘记密码,所以他决定把密码写在一张纸上.他发现这样保存密码不安全,于是他决定按下述方法加密密码:他选定一个整数 X ,保证 X 不小于 0 ,且…

Codeforces 196 E. Tricky and Cleve Password

$>Codeforces \space 196\ E. Tricky\ and\ Cleve\ Password<$ 题目大意 : 给出一个有 $n$ 个结点,$m$ 条边的连通带权无向图,有k个点设有传送门.开启的传送门可以花费 $0$ 的代价传送,走一条边要话费等同边权的代价, 一开始,所有传送门关闭, 每当你到达一个有传送门的点,那个传送门就会永久开启, 求从 $1$ 号点出发开启所有传送门所需的最小代价 $1 \leq n, m \leq 10^5$ 解题思路…

Codeforces 30 E. Tricky and Cleve Password

$>Codeforces \space 30\ E. Tricky\ and\ Cleve\ Password<$ 题目大意 : 给出一个串 $S$,让你找出 $A, B, C$ 三个串,满足 $C$ 是一个后缀, $A + B + C$是一个回文串,$B$ 是一个长度为奇数的回文串,且 $A, C$ 可以为空,并最大化 $|A| + |B| + |C|$ $1 \leq |S| \leq 10^5$ 解题思路 : 考虑 $B$ 是不能为空的,不妨先用…

CodeForces - 598A Tricky Sum （数学，快速幂的运用）

传送门: http://codeforces.com/problemset/problem/598/A A. Tricky Sum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output In this problem you are to calculate the sum of all integers from 1 to n, b…

Codeforces Round #456 (Div. 2) A. Tricky Alchemy

传送门:http://codeforces.com/contest/912/problem/A A. Tricky Alchemy time limit per test1 second memory limit per test256 megabytes Problem Description During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already…

Codeforces 429D Tricky Function（平面最近点对）

题目链接 Tricky Function $f(i, j) = (i - j)^{2} + (s[i] - s[j])^{2}$ 把$(i, s[i])$塞到平面直角坐标系里,于是转化成了平面最近点对问题. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >=…

Codeforces(429D - Tricky Function)近期点对问题

D. Tricky Function time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important cont…

[AngularJS] Sane, scalable Angular apps are tricky, but not impossible.

Read fromhttps://medium.com/@bluepnume/sane-scalable-angular-apps-are-tricky-but-not-impossible-lessons-learned-from-paypal-checkout-c5320558d4ef ng-Controller is bad, use directive instead. Don't specfiy controller in Route, use directive in templat…

const中的一些tricky的地方

1. 为了逻辑上的优化需要,const成员函数可能想修改某些成员变量,把这些成员变量定义为mutable可以绕过const的检查 2. 调用const和non-const的参数的函数可以重载 3. static_cast<const T&>(a) 将一个非const 对象转成一个const 对象 4. const_cast<T&> 将一个const对象转成非const 对象,const_cast只能放置T& 或T*…

codeforces 598A Tricky Sum

题目链接:http://codeforces.com/contest/598/problem/A 题目分类:大数题意:1到n 如果是2的次方则减去这个数,否则就加上这个数,求最后的结果是多少题目分析:数很大,关键是精度问题,刚开始用__int64和double发现都是不对的,后来发现用long long 可以过代码: #include<bits/stdc++.h> using namespace std; #define LL long long LL a[]={,,,,,,,,, ,,…

[PHP] 09 - PHP 7 & Tricky

新特征列表: 序号内容 1 PHP 标量类型与返回值类型声明 2 PHP NULL 合并运算符 3 PHP 太空船运算符(组合比较符) 4 PHP 常量数组 5 PHP 匿名类 6 PHP Closure::call() 7 PHP 过滤 unserialize() 8 PHP IntlChar() 9 PHP CSPRNG 10 PHP 7 异常 11 PHP 7 use 语句 12 PHP 7 错误处理 13 PHP intdiv() 函数 14 PHP 7 Session 选项 15 P…

[JS] Topic - define "class" by tricky methods

Ref:Javascript定义类(class)的三种方法 Javascript是一种基于对象(object-based)的语言,你遇到的所有东西几乎都是对象.但是,它又不是一种真正的面向对象编程(OOP)语言,因为它的语法中没有class(类). 那么,如果我们要把"属性"(property)和"方法"(method),封装成一个对象,甚至要从原型对象生成一个实例对象,我们应该怎么做呢? 程序员们做了很多探索,研究如何用Javascript模拟"类&qu…

codeforce 429D. Tricky Function （思维暴力过）

题目描述 Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of th…

算法笔记_055:蓝桥杯练习 Tricky and Clever Password （Java）

目录 1 问题描述 2 解决方案 1 问题描述问题描述在年轻的时候,我们故事中的英雄——国王 Copa——他的私人数据并不是完全安全地隐蔽.对他来说是,这不可接受的.因此,他发明了一种密码,好记又难以破解.后来,他才知道这种密码是一个长度为奇数的回文串. Copa 害怕忘记密码,所以他决定把密码写在一张纸上.他发现这样保存密码不安全,于是他决定按下述方法加密密码:他选定一个整数 X ,保证 X 不小于 0 ,且 2X 严格小于串长度.然后他把密码分成 3 段,最前面的 X 个字符为一段,…

问题：docker pull 用户登陆tricky，Error response from daemon: unauthorized: incorrect username or password

问题描述: PS C:\WINDOWS\system32> docker pull rabbitmqUsing default tag: latest Please login prior to pull:Login with your Docker ID to push and pull images from Docker Hub. If you don't have a Docker ID, head over to https://hub.docker.com to create one…

[CF912A]Tricky Alchemy

题意:你有a个黄水晶和b个蓝水晶,要求要x个黄水晶球(2黄),y个绿水晶球(1黄1蓝),z个蓝水晶球(3蓝),问还要多少水晶题解:模拟 C++ Code: #include<cstdio> using namespace std; long long y,b,ny,ng,nb,ans=0; long long max(long long a,long long b){return a>b?a:b;} int main(){ scanf("%I64d%I64d",&am…

[CF418E]Tricky Password

题意:有一个无限行$n$列的数表$a_{i,j}$,对于第$i\geq2$行,$a_{i,j}$为$a_{i-1,j}$在$a_{i-1,1\cdots j}$中出现的次数,要维护这个数表,支持修改第一行,查询任意位置这题挺神的...首先随机一些数据,打个表可以发现这个数表的第$2,4,6,\cdots$行都是一样的,并且$3,5,7,\cdots$行也是一样的,下面写一个来自zjt的证明(%%%) 假设对数列$a_{1\cdots n}$有变换$f(a)$:将每一个$a_i$替换成它在$a_…

Tricky Sum

In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum. For example, for n = 4 the sum is equal to - 1 - 2 + 3 - 4 = - 4, because 1, 2 and 4 are 20, 21 and 22 respective…

【Codeforces 429D】 Tricky Function

[题目链接] http://codeforces.com/problemset/problem/429/D [算法] 令Si = A1 + A2 + ... + Ai(A的前缀和) 则g(i,j) = Sj - Si f(i,j) = (i-j)^2 + (Si - Sj)^2 观察这个式子,我们发现可以用类似于平面最近点对的算法来求解该问题 [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 100010 const…

【codeforces 429D】Tricky Function

[题目链接]:http://codeforces.com/problemset/problem/429/D [题意] 给你n个数字; 让你求出一段区间[l,r] 使得 (r−l)2+(∑rl+1a[i])2最小 [题解] 求出前缀和数组sum[i]; 可以发现,如果把数组的下标i作为第一维坐标(x),前缀和sum[i]作为第二维坐标(y); 所求的式子就是任意两点之间的距离平方; 问题转化成:已知平面上的n个点; 求最近的两个点之间的距离的平方; 这个可以用分治的方法搞出来; (感觉就是个剪枝的…

Codeforces 429D Tricky Function 近期点对

题目链接:点击打开链接暴力出奇迹. 正解应该是近期点对.以i点为x轴,sum[i](前缀和)为y轴,求随意两点间的距离. 先来个科学的暴力代码: #include<stdio.h> #include<string.h> #include<vector> #include<algorithm> #include<iostream> #include<queue> using namespace std; #define N 10005…

【Codeforces Round #456 (Div. 2) A】Tricky Alchemy

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 统计需要的个数. 不够了,就买. [代码] #include <bits/stdc++.h> #define ll long long using namespace std; ll a,b,x,y,z,ta,tb; int main(){ #ifdef LOCAL_DEFINE freopen("rush_in.txt", "r", stdin); #endif ios::sync_w…

Searching with regular sentences will only get you so far – if you need to find something a bit tricky turn to these advanced yet simple methods--转

原文地址:http://www.theguardian.com/technology/2016/jan/15/how-to-use-search-like-a-pro-10-tips-and-tricks-for-google-and-beyond Search engines are pretty good at finding what you’re looking for these days, but sometimes they still come up short. For tho…

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