HDU 1395 2^x mod n = 1

PyQtWebEngine安装避坑指南:快速解决ModuleNotFoundError 本文详细解析了PyQt5开发中常见的ModuleNotFoundError: No module named 'PyQt5.QtWebEngineWidgets'错误。核心原因是PyQt5 5.11版本后,网页引擎组件被独立为PyQtWebEngine包。文章对比了退回旧版本、独立安装及使用Conda等多种解决方案,并提供了针对Windows、macOS和Linux不同平台的实战安装指南与验证方法,帮助开发者快速解决依赖问题。 阅读详情

又是一道数学题

分析:

1、n==1或者n%2==0,都不会有这样的2的幂次存在。

因为2^k(k=1、2、3...)为偶数,n为偶数时显然不存在;n==1则容易验证。

2、n为奇数是则一定存在。

n为奇数,则至少会存在一个偶数模取n等于1。2^k则会找到所有的偶数。

3、2^k%n=(2^a*2^b)%n=((2^a%n)*(2^b%n))%n。(a+b=k)

 

代码:

#include<iostream>
using namespace std;

int main()
{
    int n;
    
    while(scanf("%d",&n)!=EOF)
    {
        if(n==1 || n%2==0)
        {
            printf("2^? mod %d = 1\n",n);
        }
        else
        {
            int mi=1,tmp=2;
            
            while(tmp!=1)
            { 
                tmp=tmp*2%n;
                mi++;
            }
            
            printf("2^%d mod %d = 1\n",mi,n);
        }
    }
    
    system("pause");
    return 0;
} 


 

 

 

 

 

LeRobot 入门教程(五)异步推理 本文介绍了基于SmolVLA的异步推理技术在机器人控制中的应用。通过将动作预测与执行解耦,异步推理消除了传统同步推理中的等待延迟,使机器人行为更加流畅。文章详细讲解了如何安装依赖、启动策略服务器和机器人客户端,并重点说明了关键参数actions_per_chunk和chunk_size_threshold的调整方法。实验表明,异步推理能显著提升机器人响应速度,特别适合大规模模型应用。最后,文章建议用户根据具体硬件和任务需求优化参数配置,以获得最佳性能。 阅读详情

相关推荐

单片机按键识别篇---单击---双击----长按

  最近做一任务需要使用一个按键实现三种功能,分别是按键单击功能,按键双击功能和按键长按功能,可能之前没有接触过这类按键复用情况,顶多也只是简单识别单击和长按,没有想过双击以上按键功能,也是绞尽脑汁,想了半天,又上网看了些例程,算是对于按键的识别有点小小的了解,感觉这个功能挺实用,在此做个随笔。 一、思路   1.识别长按操作思路:我们使用系统定时器定时,然后让系统每过一段时间去扫描按键I/O...

weixin_30311605的博客 3660

HDU-1395 2^x mod n=1

See the article on https://dyingdown.github.io/2019/12/16/HDU-1395%202%5Ex-mod-n=1/ HDU-1395 2x mod n=1 Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1. Input One positive in...

yaoo_o的博客 179

【智能教育实践】MOOPer数据集在个性化学习路径规划中的应用

本文探讨了如何利用MOOPer数据集实现从“千人一面”到“千人千面”的个性化学习路径规划。通过分析学习者与平台的交互数据(如练习、反馈、讨论)并结合知识图谱,系统能够动态诊断学生知识状态,并为其推荐最优学习序列。文章详细介绍了基于数据驱动的推荐系统核心技术,包括知识追踪、图谱关联挖掘及排序策略,旨在解决传统在线教育的僵化问题,提升学习效率与动力。

read5的博客 230

hdu 1395 2^x mod n = 1

题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=1395 题目描述: 2^x mod n = 1 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12685    Ac

心之所向,无惧无悔! 871

STM32F030C8烧录实战:用J-Link Commander实现一键批量烧写(附完整脚本)

本文详细介绍了如何使用J-Link Commander工具链实现STM32F030C8的一键批量烧录,包括环境配置、批处理脚本设计、命令脚本解析以及生产环境实战技巧。通过自动化烧录流程,显著提升效率并降低人为错误,特别适合量产和频繁迭代的开发场景。

weixin_30505225的博客 462

多角度模板匹配的模板创建技巧

基于形状的模板匹配多角度创建技巧

manuoo的专栏 3403

hdu-1395 2^x mod n = 1

2^x mod n = 1 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11542 Accepted Submission(s): 3577 Problem Description Give a number n, find

有一种力量叫乐观 896

17、自适应和学习索引:在线物理设计调整的新维度

本文探讨了现代数据库管理系统中自适应索引和数据库裂解的新方法,通过结合学习模型与多臂老虎机(MAB)框架,实现在线物理设计调整。文章详细介绍了自适应索引的工作原理、数据库裂解技术以及如何利用机器学习进行索引选择和优化,并讨论了MAB框架在动态调整索引结构中的应用及其实际案例。

go5gopher的博客 82

hdu1395-2^x mod n = 1

http://acm.hdu.edu.cn/showproblem.php?pid=1395 原理为 a ^ b % n == d ;  >>>>>>  (( a % n ) ×(a % n ) ×........*(a % n ))%n == d   然后该题当n == 1 或者 n % 2 == 0 时 ,d肯定为 0 ,所以此时无解; 而当n为其他值时,必有1~n - 1的余数存在,因

bo-jwolf的专栏 1161

1.1什么是Zephyr?与FreeRTOSRT-Thread对比

我自己的项目里,三个系统都用过。这个芯片的原厂BSP哪个OS做得最靠谱?(实测,别看广告)项目核心需求是不是某个OS的“长板”?(别为了10%的功能牺牲90%的稳定性)团队三个月后能不能愉快地维护代码?(开发体验影响长期生产力)最后送一句干货:选型定不下来时,拿一块项目用的开发板,分别用三个OS跑同一个demo(比如按键控制LED+串口打印)。移植过程遇到的问题、代码看起来舒不舒服,你的直觉会告诉你答案。嵌入式没有完美解,只有适合当下的解。选好了,就踏实地写代码;

m0_50546716的博客 657

基于爬虫+词云图+Kmeans聚类+LDA主题分析+社会网络语义分析对大唐不夜城用户评论进行分析

本项目是基于携程网中关于大唐不夜城评论的文本分析,项目中用到了Python爬虫、词频分析、词云图分析、kmeans聚类、LDA主题分析、情感分析、社会网络语义分析等。本次实验我们使用了爬虫、词云图、情感分析、LDA主题分析、TF-IDF+Kmeans、社会网络语义分析等对大唐不夜城的评论进行了分析。关于博文中的源码大家可以关注派森小木屋公众号进入粉丝群领取,如对文章有疑惑,请评论区留言。

m0_64336780的博客 1万+

hdu1395 2^x mod n = 1

题意是给定一个nnn,问是否存在一个xxx使得2x mod n=12^x \ mod \ n = 12x mod n=1。 思路:首先如果n为偶数的时候肯定不存在解,所以如果存在解的话,(2,n)=1(2,n)=1(2,n)=1,一个比较好的思路是,由于由于2是原根,所以这个题目就是求2 模 7的阶,根据原根的性质,我们只需要在φ(n)的因子中用快速幂找到最...

一身清贫怎敢入繁华 381

OV9734 DATASHEET,完整规格书

OV9734完整规格书,包括电气参数,尺寸规格,寄存器配置表等

HDU_1395 2^x mod n = 1

2^x mod n = 1

yuniZZ的专栏 611

深度学习工具包 DeepLearningToolbox

深度学习工具包Deprecation notice.-----This toolbox is outdated and no longer maintained.There are much better tools available for deep learning than this toolbox, e.g. [Theano](http://deeplearning.net/software/theano/), [torch](http://torch.ch/) or [tensorflow](http://www.tensorflow.org/)I would suggest you use one of the tools mentioned above rather than use this toolbox.Best, Rasmus.DeepLearnToolbox================A Matlab toolbox for Deep Learning.Deep Learning is a new subfield of machine learning that focuses on learning deep hierarchical models of data.It is inspired by the human brain's apparent deep (layered, hierarchical) architecture.A good overview of the theory of Deep Learning theory is[Learning Deep Architectures for AI](http://www.iro.umontreal.ca/~bengioy/papers/ftml_book.pdf)For a more informal introduction, see the following videos by Geoffrey Hinton and Andrew Ng.* [The Next Generation of Neural Networks](http://www.youtube.com/watch?v=AyzOUbkUf3M) (Hinton, 2007)* [Recent Developments in Deep Learning](http://www.youtube.com/watch?v=VdIURAu1-aU) (Hinton, 2010)* [Unsupervised Feature Learning and Deep Learning](http://www.youtube.com/watch?v=ZmNOAtZIgIk) (Ng, 2011)If you use this toolbox in your research please cite [Prediction as a candidate for learning deep hierarchical models of data](http://www2.imm.dtu.dk/pubdb/views/publication_details.php?id=6284)```@MASTERSTHESIS\{IMM2012-06284, author = "R. B. Palm", title = "Prediction as a candidate for learning deep hierarchical models of data", year = "2012",}```Contact: rasmusbergpalm at gmail dot comDirectories included in the toolbox-----------------------------------`NN/` - A library for Feedforward Backpropagation Neural Networks`CNN/` - A library for Convolutional Neural Networks`DBN/` - A library for Deep Belief Networks`SAE/` - A library for Stacked Auto-Encoders`CAE/` - A library for Convolutional Auto-Encoders`util/` - Utility functions used by the libraries`data/` - Data used by the examples`tests/` - unit tests to verify toolbox is workingFor references on each library check REFS.mdSetup-----1. Download.2. addpath(genpath('DeepLearnToolbox'));Example: Deep Belief Network---------------------```matlabfunction test_example_DBNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit RBM and visualize its weightsrand('state',0)dbn.sizes = [100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);figure; visualize(dbn.rbm{1}.W'); % Visualize the RBM weights%% ex2 train a 100-100 hidden unit DBN and use its weights to initialize a NNrand('state',0)%train dbndbn.sizes = [100 100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);%unfold dbn to nnnn = dbnunfoldtonn(dbn, 10);nn.activation_function = 'sigm';%train nnopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.10, 'Too big error');```Example: Stacked Auto-Encoders---------------------```matlabfunction test_example_SAEload mnist_uint8;train_x = double(train_x)/255;test_x = double(test_x)/255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit SDAE and use it to initialize a FFNN% Setup and train a stacked denoising autoencoder (SDAE)rand('state',0)sae = saesetup([784 100]);sae.ae{1}.activation_function = 'sigm';sae.ae{1}.learningRate = 1;sae.ae{1}.inputZeroMaskedFraction = 0.5;opts.numepochs = 1;opts.batchsize = 100;sae = saetrain(sae, train_x, opts);visualize(sae.ae{1}.W{1}(:,2:end)')% Use the SDAE to initialize a FFNNnn = nnsetup([784 100 10]);nn.activation_function = 'sigm';nn.learningRate = 1;nn.W{1} = sae.ae{1}.W{1};% Train the FFNNopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.16, 'Too big error');```Example: Convolutional Neural Nets---------------------```matlabfunction test_example_CNNload mnist_uint8;train_x = double(reshape(train_x',28,28,60000))/255;test_x = double(reshape(test_x',28,28,10000))/255;train_y = double(train_y');test_y = double(test_y');%% ex1 Train a 6c-2s-12c-2s Convolutional neural network %will run 1 epoch in about 200 second and get around 11% error. %With 100 epochs you'll get around 1.2% errorrand('state',0)cnn.layers = { struct('type', 'i') %input layer struct('type', 'c', 'outputmaps', 6, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %sub sampling layer struct('type', 'c', 'outputmaps', 12, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %subsampling layer};cnn = cnnsetup(cnn, train_x, train_y);opts.alpha = 1;opts.batchsize = 50;opts.numepochs = 1;cnn = cnntrain(cnn, train_x, train_y, opts);[er, bad] = cnntest(cnn, test_x, test_y);%plot mean squared errorfigure; plot(cnn.rL);assert(er<0.12, 'Too big error');```Example: Neural Networks---------------------```matlabfunction test_example_NNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);% normalize[train_x, mu, sigma] = zscore(train_x);test_x = normalize(test_x, mu, sigma);%% ex1 vanilla neural netrand('state',0)nn = nnsetup([784 100 10]);opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samples[nn, L] = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.08, 'Too big error');%% ex2 neural net with L2 weight decayrand('state',0)nn = nnsetup([784 100 10]);nn.weightPenaltyL2 = 1e-4; % L2 weight decayopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex3 neural net with dropoutrand('state',0)nn = nnsetup([784 100 10]);nn.dropoutFraction = 0.5; % Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

ssm框架搭建

这个是一个在idea上成功跑起来的ssm框架项目,保存在这里方便以后使用直接下载

HDU 1395 2^x mod n = 1

Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1. Input One positive integer on each line, the value of n. Output If the minimum x exists, print a line with 2^x mod n = 1. Prin...

weixin_44705665的博客 111

迈瑞BS系列全自动生化分析仪LIS协议接口手册(内含HL7和ASTM)

迈瑞BS系列全自动生化分析仪LIS协议接口手册,内含HL7和ASTM)

hdu 1395 2^x mod n = 1 暴力 欧拉函数

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1395 问题描述 给出一个数字n,找到满足2 ^ x mod n = 1的最小x(x> 0)。 输入 每行一个正整数,n的值。 产量 如果存在最小x,则打印一条2 ^ x mod n = 1的行。 打印2 ^ ?mod n = 1否则。 您应该用特定数字替换x和n。 样本输入 ...

一叶之修的博客 557

别再用‘围追堵截’了!从PCB布线到软件展频,聊聊EMC整改中那些真正省钱的‘内功心法’

本文深入探讨了EMC整改中的高效降本策略,从PCB布线到软件展频技术,揭示了如何在产品设计初期构建EMC免疫体系。通过关键器件选型、电源网络净化和接地系统优化等实践方法,显著降低后期整改成本,提升产品可靠性。特别推荐内置展频功能的时钟发生器等经济高效的解决方案。

weixin_33743880的博客 336

(HDU)1395 -- 2^x mod n = 1

题目链接:http://vjudge.net/problem/HDU-1395 分析可知,n为偶数或者1的时候输出‘?’ 自己设置一个循环周期。 1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <iostream> 5 ...

diupouhao4253的博客 192

HDU 1395 2^x mod n = 1 数论

题目地址:  http://acm.hdu.edu.cn/showproblem.php?pid=1395 这里可以有一个结论,大于等于1的奇数都有2^t mod n=1 代码如下: #include #include #include #include #include #include #include #include #include

lamborghini1993 1164

HDU 1395 2^x mod n = 1

2^x mod n = 1 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6825    Accepted Submission(s): 2049 Problem Description Give a number

闲暇时光写代码 1673

HDU 1395 2^x mod n = 1

/* 中文题意: 中文翻译: 题目大意:求出最小的 n 使得2的 I 次方对 n 的值为1. 解题思路:如下: 难点详解:先用费马小定理了解2的 i 次方对偶数取余都不可能是一,还有就是排除 1 。之后要用中国剩余定理让 t 的值不超出 int 范围。不用这个定理我错了n次,都是超时。我猜测可能是 t 的值超出了int 的范围了,之后的数都是随机的,所以一直运行不出来,才会超时的。(不知

淡定的小鱼 779

hdu 1395 2^x mod n = 1(欧拉函数)

2^x mod n = 1 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16236    Accepted Submission(s): 5053 Problem Description Give a numb

TUT我好菜啊 1169

【HDU】 1395 2^x mod n = 1

2^x mod n = 1题目链接 2^x mod n = 1 题目大意    找到一个最小的x满足2x mod n = 12^x\ mod\ n\ =\ 1如果没有则输出2? mod n = 12^?\ mod\ n\ =\ 1题解    我们可以把式子化为(2∗2x−1) mod n = 1(2*2^{x-1})\ mod\ n\ =\ 1 首先可以看到2x2^x是个偶数,要想让该式成立肯定n不

S_Black的专栏 461
上一篇: 1/8/2012 ICPC培训 第十七天
下一篇: 真想说再见,隐身帝们
lulipeng_cpp
博客等级 码龄15年 169粉丝 130原创
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付:10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值