River Hopscotch [二分]

android usb读写,安卓(Android)下如何开发USB NFC读写器app 对安卓工程师来说,在安卓下使用USB设备需要了解很多硬件的内容,这可能会导致工程周期的延长或者app的不稳定。为了将这种风险降到最低,友我科技发布了NFC读写器在安卓下的sdk,使用NFC读写器的类接口,安卓工程师就可以直接调用NFC读写器了,简单可靠。NFC读写器的安卓库文件为:yoworfidreader.jar在ADT工程中导入yoworfidreader.jar库文件,即可在工程中引用NF... 阅读详情

题意:马跳石头,给定石头的位置,求拿掉m个石头后,马到目的地过程中跳跃的最小距离的最大值。
思路:给定的石头位置,初始0,目的地L, 排个序。然后二分最小距离找最大值。
二分的单调性:如果马的最小一步能跨越l,那么对于任意的li>l作为最小步长都是合理的。
二分的check(mid):如果当前最小步长mid满足仅有cnt<=m个小于mid的长度时,以当前mid作为最小步长是可行的。

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<cstdlib>
#include<list>
#include<stack>
#include<cmath>
#include<iomanip>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
void debug() {cout << "ok running!" << endl;}
int a[100005];
int m, n, L;
bool check(int mid)
{
    int cnt = 0;
    for(int i = 0; i <= n; ++i)
    {
        int temp = a[i];
        while(a[i+1] - temp < mid && i <= n)
        {
            cnt++;
            i++;
        }
    }
    if(cnt <= m)
    {
        return 1;
    }
    else return 0;
}
int main()
{
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    while(cin >> L >> n >> m)
    {
        a[0] = 0;
        a[n+1] = L;
        for(int i = 1; i <= n; ++i)
            cin >> a[i];
        sort(a, a+n+1);
        int l = 0, r = L;
        int ans = -1;
        while(l <= r)
        {
            int mid = (l+r) >> 1;
            if(check(mid))
            {
                l = mid+1;
                ans = mid;
            }

            else r = mid-1;
        }
        cout << ans << endl;
    }
    return 0;
}
数据治理路径之辩:从“先治后用”到“边用边治”,企业如何选择最优路径? 数据治理是企业在数智化建设中的核心挑战,主要涉及三种时序策略:先治理后使用、先用后治理和边用边治。先治理后使用强调在数据应用前建立完整的治理框架,适合高合规要求的行业,但可能响应滞后。先用后治理则优先释放数据价值,适合快速变化的业务环境,但可能累积数据质量问题。边用边治寻求动态平衡,通过持续改进实现治理与应用的同步迭代,适合复杂数据环境。技术选型决策矩阵和实施方法论提供了具体的操作指南,建议根据组织特性和业务需求选择适配方案,并通过渐进式演进路径实现治理与业务的共生演进。 阅读详情

相关推荐

OpenCV实战:用C++和Python实现亚像素边缘检测(附完整代码)

本文详细介绍了使用OpenCV实现亚像素边缘检测的技术,通过C++和Python双语言代码解析,提升工业测量和医学影像中的检测精度。文章涵盖算法原理、核心实现步骤及工业应用案例,帮助开发者掌握亚像素精度处理技术,解决高精度场景中的边缘检测难题。

beer8的博客 373

River Hopscotch(二分)

题目链接:River Hopscotch 题意:小牛过河起点到终点距离为L,有m块石头,可以移除M块石头,寻找所跳最小距离的最大值 题解:枚举所跳最小的距离的最大值,若两个石头之间的实际距离小于等于答案距离,则这块石头需要搬走 #include<stdio.h> #include<algorithm> using namespace std; int a[1055000]; int main() { int L,n,m; int i,j; scanf("%d%d%d",&amp

weixin_52513845的博客 303

SIMetrix教程-001.SIMetrix软件简介与安装

  由于某些原因需要用到SIMetrix仿真软件,然而网上的资料并不是特别多,故在此记录一下这款仿真软件的学习过程,也给有需要的人提供一些参考,免走弯路。如果使用过Pspice或者LTspice软件,学习SIMetrix会很容易入门,仿真软件都是大同小异的,会一样后都可以快速入门其他软件。   SIMetrix是一款以SPICE模型为基础的仿真软件,集成了SIMetrix和SIMPLIS两款内核,为工程师提供了可靠的、易于使用的、并完美结合了速度与精度的电路仿真方式。官方的话不多说,和LTspice一样,优

gaoyong_wang的博客 2万+

River Hopscotch(sort+二分法)

G - River Hopscotch Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Description Every year the cows hold an event featuring a peculiar versio

唐唐 463

ORCAD原理图关联封装时出现ERROR(ORCIS-6245)、ERROR(ORCIS-6013)问题

1、当你进行如下操作时: 系统先是提示你出现ERROR(ORCIS-6245)然后再确定后再出现ERROR(ORCIS-6013)问题 是由于你的数据库的路径不对。 具体操作方法如下:1、先找到你的CAPTURE.INI文件。 2、用记事本打开CAPTURE.INI。找到下图路径 3、在你的Orcad capture CIS原理图软件中点击 Options------>CIS con...

weixin_42304393的博客 2万+

半监督的医学分割数据集(LA, Pancreas, ACDC和PROMISE12)分享

这篇文章提供了四个常用的半监督医学分割数据集,所有数据集都已经被处理成了.h5文件。下载之后就可以直接使用

qq_45691955的博客 4171

二分PkU3258

<span style="color:#330099;">/* E - 二分 Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Description Every year the cows hold an event featurin...

weixin_30538029的博客 195

昇腾,Ascend,NPU,mindie,镜像,部署vllm:第5篇,昇腾环境下部署蒸馏版deepseek-r1,并在dify中注册模型供应商

这两天没少折腾deepseek-r1。之前在A100上用ollama部署了一份DeepSeek-R1-Distill-Qwen-32B,感觉真的特别慢。现在,再用昇腾部署一遍,看看速度能不能更快。p.s. 类似的内容其实已经写了很多遍了。但感觉之前写得都太碎,不够顺畅。接着新模型写篇长文,把能包括的都包括进来。这样大家也不需要到处翻来翻去。

xiezhipu的博客 2450

River Hopscotch (二分+贪心)

River Hopscotch Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straigh...

YOONGI 320

倾斜摄影数据转换为3DTile格式的方法:基于CesiumLab GIS

通过使用CesiumLab GIS,倾斜摄影数据可以方便地转换为3DTile格式,并在CesiumJS等3D地理信息可视化工具中进行展示和分析。一旦成功导入3DTile数据,就可以在CesiumLab GIS中利用其强大的可视化和分析功能进行进一步的处理。转换完成后,可以将生成的3DTile数据导入到CesiumLab GIS中进行可视化和分析。启动CesiumLab GIS应用程序,并按照界面上的指导导入转换后的3DTile数据。脚本将开始执行,并将倾斜摄影数据转换为3DTile格式。

ObDjango的博客 1075

CMM软件:PC-DMIS汽车制造二次开发

PC-DMIS 是一种广泛应用于坐标测量机(CMM)的测量软件,它不仅提供了强大的测量功能,还支持用户通过二次开发来扩展其功能。二次开发可以显著提升测量效率,满足特定的测量需求,减少重复性工作,提高数据处理和分析的准确性。本节将介绍 PC-DMIS 二次开发的基本概念、开发环境的搭建以及开发中常用的工具和方法。PC-DMIS 二次开发为用户提供了强大的工具,可以显著提升测量效率、减少重复性工作并提高数据处理和分析的准确性。

kkchenjj的博客 1813

poj3258 二分及一些思考

题意:牛要到河对岸,在与河岸垂直的一条线上,河中有N块石头,给定河岸宽度L,以及每一块石头离牛所在河岸的距离,           现在去掉M块石头,要求去掉M块石头后,剩下的石头之间以及石头与河岸的最小距离的最大值。 用二分做,但是开始写了三个版本的二分,全都wa。 无赖看了别人的二分,还是不理解,为什么他们写的就能过。 反复思索后,终于明白了:关键在于题目求的是什么。 做题思想:二分

5383

PNETLab 镜像包官方下载太慢?不急,最新版本PNET_4.2.10分享!

你好,这里是网络技术联盟站,我是瑞哥。PNETLab是一个强大的网络仿真工具平台,全称Packet Network Emulator Tool Lab。它不仅允许用户进行深入的网络实验,还支持社区成员之间的实验室共享。PNETLab致力于为学习者、教师以及网络工程师提供一个全面的实验环境,以便于他们能够在实际部署前测试和验证网络配置和协议。和。这两个部分相辅相成,共同构成了一个全面且灵活的网络仿真生态系统。

网络技术联盟站 4466

POJ 3258 River Hopscotch 二分

来源:http://poj.org/problem?id=3258 题意:有一条河,河的长度已知,河中间有一些石头,石头的数量知道,相邻两块石头之间的距离已知。现在可以移除一些石头,问移除m块石头后,相邻两块石头之间的距离的最小值最大是多少。 思路:二分枚举答案。每 次二分枚举一个值,判断该值能够去掉多少块石头。二分枚举求上限。 代码: #include #include #inclu

3040

Statistics and Machine Learning Toolbox

Statistics and Machine Learning Toolbox(带书签),MATLAB统计学习工具

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")

POJ - 3258 River Hopscotch(二分找最小值最大)

POJ - 3258 River Hopscotch(二分找最小值最大) Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes plac...

weixin_43179892的博客 199

jre9下载 jdk9

(Java SE Runtime Environment 9.0) 9.0u175 官方最新版 【x86|x64】

POJ 3258 River Hopscotch 二分

POJ 3258 River Hopscotch 二分 题意:移石头,两个岩石之间的距离为D,从一个岩石出发跳到另一个岩石, 找最大化的最短距离。   从一个石头i开始,跳到岩石j,i到j之间距离如果大于mid,则把i与j之间的(j-i-1)个石头全部移走 否则,j继续++   如果移走的石头小于等于m,证明mid太小了,需要大一点的mid 如果移走的石头大于m,证明mid太大了...

LMengi000的博客 201

Camera2Demo

在Camera2 的基础上进行封装,使得Camera2一堆初始化逻辑与Activity分离,降低耦合度,完成预览拍照并保存的操作

River Hopscotch【二分】

River Hopscotch  POJ - 3258  Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a...

C++的星空 208

River Hopscotch 二分

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a ro...

博客标题不能为空我也没办法 229

3258 River Hopscotch 二分答案

River HopscotchTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 2432 Accepted: 1063<br />Description<br />Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The

kongming_acm的专栏 706

POJ 3258 River Hopscotch 二分 最小值最大化

POJ 3258 River Hopscotch 奶牛过河,起点为0,终点到起点的距离为 L ,中间有n块石头,要去掉m块,求去掉m块后 使奶牛至少要跳的最小距离为所有方法中的最大值,并输出这个值(这道题是要求一堆最小距离里面的最大值) 思路 使用二分法。这道题里面注意牛能跳的距离是没有限制的,就是说牛想跳多远就跳多远,对于牛最多能跳多少次也没有限制!但是要求里面的最小值,所以这道题二分法所...

没有梦想何以远行 226

poj 3258 River Hopscotch 【二分】

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9859   Accepted: 4211 Description Every year the cows hold an event featuring a peculiar ve

世界很大 688

POJ3258 River Hopscotch(二分)

River Hopscotch 二分(判定函数有点难想~)

Todd_Dijkstra的博客 386

从‘跳房子’到‘二分答案’:手把手教你用C++解决USACO经典题River Hopscotch(附洛谷P2855 AC代码)

本文详细介绍了如何使用C++和二分答案算法解决USACO经典题目River Hopscotch,包括题目理解、算法框架搭建、check函数设计以及多平台AC代码实现。通过游戏化类比和实战技巧,帮助读者掌握二分答案算法的核心思想与应用,适用于信息学奥赛和算法竞赛准备。

weixin_30851867的博客 378

River Hopscotch 二分查找

Description  Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight ri...

我是谁 --- 卡比兽 203

POJ 3258:River Hopscotch 二分的好想法

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9326   Accepted: 4016 Description Every year the cows hold an event featuring a peculiar ve

2622

River Hopscotch 二分

River HopscotchTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 8975 Accepted: 3859DescriptionEvery year the cows hold an event featuring a peculiar version of hopscotch that involves careful...

yihanyifan的博客 197
上一篇: Dying[二分]
下一篇: Painting Fence[分治]
Qer_
博客等级 码龄11年 34粉丝 69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