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 equivale…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3609 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 e…

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 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 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 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…

Description 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 toax≡1 (mod m). Input There are multiple test cases. The first line of input is an integer T ≍ 2000 indicating…

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…

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…

The modular modular multiplicative inverse of an integer a modulo m is an integer xsuch 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…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4712 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.…

题 题意 求a关于m的乘法逆元 分析 a x ≡ 1 (mod m) 等价于 ax+my=1 求x的最小正数(不能是0,我就WA在这里了). 当m=1时,或者 gcd(a,m)!=1 时x不存在. 所以用扩展gcd就可以求了. 代码 #include<cstdio> #define ll long long ll exgcd(ll a,ll b,ll &x,ll &y) { if(b==0) { x=1; y=0; return a; } ll r=exgcd(b,a%b,y,…

Alice bought a lot of pairs of socks yesterday. But when she went home, she found that she has lost one of them. Each sock has a name which contains exactly 7 charaters. Alice wants to know which sock she has lost. Maybe you can help her. Input There…

1. 题目描述求乘法逆元. 2. 基本思路利用扩展gcd求逆元,模板题目. 3. 代码 /* 3609 */ #include <iostream> #include <sstream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <…

点我看题目 题意 : 这个题是求逆元的,怎么说呢,题目看着很别扭....就是给你a和m,让你求一个最小的x满足a-1≡x (mod m).或者ax≡1 (mod m).通俗点说呢,就是找一个最小的x,他满足的条件的是a*x取余m等于1. 思路 :这个题反正数据不是很大,枚举就行了,因为维基百科中说的,两个数必须是互质的,所以判断一下,还有这个题特别逗的是m是1的时候x肯定也是1,我觉得任何一个数取余1都应该是0,可是这里显然不是这样的.这个题还可以用扩展的欧几里德算法,也可以做. #includ…

题意:求乘法逆元最小正正数解 思路:a*x≡1(mod m),则称x 是 a 关于 m 的乘法逆元,可以通过解a*x + m*y = 1解得x.那么通过EXGcd得到特解x1,最小正解x1 = x1 % m,如果x1 <=0,x1 += m,注意m是负数时取绝对值,因为是正解,所以不能用(x1%m+m)%m. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<cmath>…

#include <bits/stdc++.h> using namespace std; template <typename T> T inverse(T a, T m) { T u = 0, v = 1; while (a != 0) { T t = m / a; m -= t * a; swap(a, m); u -= t * v; swap(u, v); } assert(m == 1); return u; } template <typename T> c…

PROBLEM LINK: PracticeContest Author: adminTester: Kevin AtienzaEditorialist: Ajay K. VermaRussian Translator: Sergey KulikMandarian Translator: Gedi Zheng DIFFICULTY: Medium PREREQUISITES: Combinatorics, Modular Arithmetic PROBLEM: Given two integer…

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…

From LB@10.0.0.55 Misc 0x01 misc100(图片隐写) 首先用binwalk扫了一下,发现没毛病. 然后就搜了一下jpg的文件尾FFD9,如下图,看到了png格式的标志IHDR. 于是将FFD9以前的部分删除,补全PNG文件头8950 4e47 0d0a 1a0a得到一张新的图片,看上去是全白的,毫无内容.但是打开提示图片出错,于是想到是宽高和CRC不匹配导致的. 分析一波png格式可以知道, CRC = 0x9A768270,width = 0x0320,heigh…

from gmpy2 import * import libnum n = 0x00b0bee5e3e9e5a7e8d00b493355c618fc8c7d7d03b82e409951c182f398dee3104580e7ba70d383ae5311475656e8a964d380cb157f48c951adfa65db0b122ca40e42fa709189b719a4f0d746e2f6069baf11cebd650f14b93c977352fd13b1eea6d6e1da775502ab…

The 9th Zhejiang Provincial Collegiate Programming Contest A    Taxi Fare    25.57% (166/649)     (水题)    计算两种计程车计费方式的差值, 注意四舍五入    B    Unrequited Love    14.00% (7/50)    水题) C    Count the Trees    9.87% (8/81)   (二叉树)    卡特兰数应用的问题     D    Draw S…

拖了一周才完成的题解,抛出一个可爱的表情 (っ'-')╮ =͟͟͞͞❤️.对我来说E.F比较难,都是线段树的题,有点久没写了. A - Infinite Sequence CodeForces - 675A 公差为c,首项为a的等差数列是否有一项等于b. 注意c为0的情况. #include<cstdio> long long a,b,c; int main() { scanf("%lld%lld%lld",&a,&b,&c); if(c==0&am…

Problem Statement You are participating in a quiz with N+M questions and Yes/No answers. It's known in advance that there are N questions with answer Yes and M questions with answer No, but the questions are given to you in random order. You have no…

RSA介绍 根据加密原理,可以将大部分的加密算法分为两大类:对称加密算法和非对称加密算法.对称加密算法的加密和解密采用的是同一套算法规则.而非对称加密算法加密时用的是公钥(公开给所有人),解密时用的是私钥(只有相关人员拥有), 非对称加密算法中使用最广泛的就是RSA算法.RSA算法非常可靠,密钥越长,就越难破解.当今互联网中已经纰漏的破解方法是针对768位密钥.所以一般认为1024位的密钥加密是安全的,2048位是绝对安全的 RSA算法原理 RSA的算法是基于一个大因数是很难计算分解这一原理的.…

链接:  https://www.codechef.com/FEB18/problems/BROCLK Broken Clock Problem Code: BROCLK Chef has a clock, but it got broken today — the minute hand on Chef's clock doesn't rotate by the angle 2π/3600 each second, but by a different fixed angle x. The c…

一.拖入ida,来静态分析F5大法好 要注意的点: 1._gmz_init_set_str() 这个函数,也是看师傅的wp,学到的,以后还是得多google, 本质上是这个函数: int mpz_init_set_str (mpz_t rop, const char *str, int base) [Function] 将str以base进制送入rop的内存 2._gmpz_pown()这个函数 void mpz_powm (mpz_t rop, const mpz_t base, const…

最近以来,我使用d3进行我的可视化工具的开发已经3个月了,同时也兼用其他一些图表类库,自我感觉稍微有点心得.之前我也写过相关文章,我涉及的数据可视化的实现技术和工具,但是那篇文章对于项目开发而言太浅了.于是想写关于d3进行项目实战的系列文章,就像我之前的angularjs实战系列文章一样把整个开发过程中遇到的各种问题及解决办法梳理成章,以为留存.作为开篇,我还是想先把这段时间来我一直参考的资料做一个整理,并谈一些宏观的体会. 一.前方有坑,注意! ————————————————————————…

HITCON 2019 Lost Modular again writeup 算是基础题,有很多之前题的影子,做不出来纯属菜. 题目 加密脚本 from Crypto.Util.number import * class Key: def __init__(self, bits): assert bits >= 512 self.p = getPrime(bits) self.q = getPrime(bits) self.n = self.p * self.q self.e = 0x10000…

上一节中我们学会了如何旋转x轴标签以及自定义标签内容,在这一节中,我们将接触动画(transition) 首先,我们要在页面上添加一个按钮,当我们点击这个按钮时,调用我们的动画.所以,我们还需要在原来的基础上添加两个东西. 添加一个按钮 <div id="option"> <input name="updateButton" type="button" value="Update" onclick="…