Problem description The following problem is well-known: given integers n and m, calculate 2n mod m, where 2n = 2·2·...·2 (n factors), and x mod y denotes the remainder of division of x by y. You are asked to solve the "reverse" problem. Given i…

题目链接 题解:这道题比赛的时候,学弟说是网络流,当时看N这么大,觉得网络流没法做,实际本题通过巧妙的建图,然后离散化. 先说下建图方式,首先每个覆盖区域,只有左右端点,如果我们只用左右端点的话,最多有400个点,所以第一步离散化.每个$i$和$i+1$连一条边流量为K,费用为0的边,每一个覆盖区域从左端点$u$向右端点$v$连一条流量为1,费用为$-w$的边,因为一般算最大费用,就要把边权变成负的算最小,最后取反是最大.对了,右端点$v$要+1,因为这个覆盖区域是从$u->v$,我应该从$v+…

很明显的区间K覆盖模型,用费用流求解.只是这题N可达1e5,需要将点离散化. 建模方式步骤: 1.对权值为w的区间[u,v],加边id(u)->id(v+1),容量为1,费用为-w; 2.对所有相邻的点加边id(i)->id(i+1),容量为正无穷,费用为0; 3.建立源点汇点,由源点s向最左侧的点加边,容量为K,费用为0,由最右侧的点向汇点加边,容量为K,费用为0 4.跑出最大流后,最小费用取绝对值就是能获得的最大权 #include<bits/stdc++.h> using n…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 当a<b的时候 a%b==a 显然2^n增长很快的. 当2^n>=1e8的时候,直接输出m就可以了 [代码] #include <bits/stdc++.h> #define ll long long using namespace std; ll n,m; int main(){ #ifdef LOCAL_DEFINE freopen("rush_in.txt", "r",…

Rsapaper.pdf http://people.csail.mit.edu/rivest/Rsapaper.pdf [概述Abstract 1.将字符串按照双方约定的规则转化为小于n的正整数m,可能分为多段,这不是关键: 2.加密过程同解密过程,都是取明/密文的public/private次方,然后对公共的n取余数: 3.整数转化为字符串 ] A message is encrypted by representing it as a number M, raising M to a pu…

https://en.wikipedia.org/wiki/RSA_(cryptosystem) RSA is one of the first practical实用性的 public-key cryptosystems and is widely used for secure data transmission. In such a cryptosystem密码系统, the encryption key is public and differs from the decryption…

Tips <Effective Java, Third Edition>一书英文版已经出版,这本书的第二版想必很多人都读过,号称Java四大名著之一,不过第二版2009年出版,到现在已经将近8年的时间,但随着Java 6,7,8,甚至9的发布,Java语言发生了深刻的变化. 在这里第一时间翻译成中文版.供大家学习分享之用. 17. 最小化可变性 不可变类简单来说是它的实例不能被修改的类. 包含在每个实例中的所有信息在对象的生命周期中是固定的,因此不会观察到任何变化. Java平台类库包含许多不…

SSL加速卡调研的原因及背景 SSL加速卡调研的原因及背景 网络信息安全已经成为电子商务和网络信息业发展的一个瓶颈,安全套接层(SSL)协议能较好地解决安全处理问题,而SSL加速器有效地提高了网络安全处理的性能. 公司产品本身可以在软件层次,利用CPU和内存等资源可以处理SSL加解密,但是性能不高.在公司产品中集成SSL加速卡,有利于提高处理SSL加解密的性能,节省系统资源,并提高公司产品的整体工作性能. 公司产品的需求: 1)         所有网络接口的数据流量都可以经过SSL加速卡的处理…

A. Magic Mirror #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <time.h> #include <string> #include <set> #include <map> #include <list> #include <stack> #includ…

A. Modular Exponentiation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The following problem is well-known: given integers n and m, calculate , where 2n = 2·2·...·2 (n factors), and  den…

https://en.wikipedia.org/wiki/Chinese_remainder_theorem 中国剩余定理 https://en.wikipedia.org/wiki/RSA_(cryptosystem) The public key is (n = 3233, e = 17). For a padded plaintext message m, the encryption function is The private key is (n = 3233, d = 413).…

A. Magic Mirror 水. #include <bits/stdc++.h> using namespace std; int t; ]; inline bool work() { int len = strlen(s); ) return false; ] != ] != 'J') return false; ] != ] != 'E') return false; ] != ] != 'S') return false; ] != ] != 'S') return false;…

General overview[edit]   Illustration of the idea behind Diffie–Hellman key exchange Diffie–Hellman key exchange establishes a shared secret between two parties that can be used for secret communication for exchanging data over a public network. The…

这场打得还是比较爽的,但是队友差一点就再过一题,还是难受啊. 每天都有新的难过 A. Magic Mirror Jessie has a magic mirror. Every morning she will ask the mirror: 'Mirror mirror tell me, who is the most beautiful girl in the world?' If the mirror says her name, she will praise the mirror: '…

https://en.wikipedia.org/wiki/Modular_exponentiation 蒙哥马利(Montgomery)幂模运算是快速计算a^b%k的一种算法,是RSA加密算法的核心之一. 蒙哥马利模乘的优点在于减少了取模的次数(在大数的条件下)以及简化了除法的复杂度(在2的k次幂的进制下除法仅需要进行左移操作).模幂运算是RSA 的核心算法,最直接地决定了RSA 算法的性能. 针对快速模幂运算这一课题,西方现代数学家提出了大量的解决方案,通常都是先将幂模运算转化为乘模运算.…

题目顺序:A F G H I K L 做题链接 A. Magic Mirror 题意:判断 给出的 字符串 是否等于"jessie",需要判断大小写 题解:1.用stl库 tolower ... 2.用ascii码将大写转换为小写,例如: 'A' + 32 = 'a' *F. Modular Production Line 题解:网络流最长区间K覆盖问题, 洛谷P3358 参考题解 参考代码 G. Give Candies 题意:n个小朋友,n个糖,每个小朋友按顺序拿随机数量个糖,直到…

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Problem F: ExponentiationTime Limit: 1 Sec Memory Limit: 128 MBSubmit: 4 Solved: 2[Submit][Status][Web Board] [Edit] [TestData]DescriptionProblems involving the computation of exact values of very large magnitude and precision are common. For example…

DescriptionProblems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems. This problem requires that you write a…

akka-stream的Graph是一种运算方案,它可能代表某种简单的线性数据流图如:Source/Flow/Sink,也可能是由更基础的流图组合而成相对复杂点的某种复合流图,而这个复合流图本身又可以被当作组件来组合更大的Graph.因为Graph只是对数据流运算的描述,所以它是可以被重复利用的.所以我们应该尽量地按照业务流程需要来设计构建Graph.在更高的功能层面上实现Graph的模块化(modular).按上回讨论,Graph又可以被描述成一种黑盒子,它的入口和出口就是Shape,而内部的…

The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1≡x (mod m). This is equivalent to ax≡1 (mod m). Input There are multiple test cases. The first line of input is an integer T ≍ 2000 indicating the number…

In data structure Hash, hash function is used to convert a string(or any other type) into an integer smaller than hash size and bigger or equal to zero. The objective of designing a hash function is to "hash" the key as unreasonable as possible.…

A modular application is an application that is divided into a set of loosely coupled functional units (named modules) that can be integrated into a larger application. A client module encapsulates a portion of the application's overall functionality…

题目来源:http://poj.org/problem?id=1060 题目大意: 考虑系数为0和1的多项式.两个多项式的加法可以通过把相应次数项的系数相加而实现.但此处我们用模2加法来计算系数之和.一个实例: (x^6 + x^4 + x^2 + x + 1) + (x^7 + x + 1) = x^7 + x^6 + x^4 + x^2 同样,多项式的减法我们在求系数之差时也用模2减法,例如: (x^6 + x^4 + x^2 + x + 1) - (x^7 + x + 1) = x^7 +…

Modular multiplication of polynomials Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3239   Accepted: 1459 Description Consider polynomials whose coefficients are 0 and 1. Addition of two polynomials is achieved by 'adding' the coeffici…

 Modular Production Line 时空限制: 1000ms /65536K   An automobile factory has a car production line. Now the market is oversupply and the production line is often shut down. To make full use of resources, the manager divides the entire production line in…

Modular multiplication of polynomials Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4377   Accepted: 1980 Description Consider polynomials whose coefficients are 0 and 1. Addition of two polynomials is achieved by 'adding' the coeffici…

Modular Inverse Time Limit: 2 Seconds      Memory Limit: 65536 KB The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1≡x (mod m). This is equivalent to ax≡1 (mod m). Input There are multiple test cases. Th…

                             Modular Fibonacci The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence:F0 = 0F1 = 1Fi = Fi−1 + Fi−2 for i > 1Write a program which calculates Mn = Fn mod 2m for given pair of n and…

[网络流]Modular Production Line 焦作上的一道,网络流24题中的原题.... https://nanti.jisuanke.com/t/31715 给出了1e5个点,但是因为最多200条边,也就是最多用到400个点,所用先离散化,然后建图. 建图: 1.对权值为w的区间[u,v],加边id(u)->id(v+1),容量为1,费用为-w; 2.对所有相邻的点加边id(i)->id(i+1),容量为正无穷,费用为0; 3.建立源点汇点,由源点s向最左侧的点加边,容量为K,费…