HDU5734 Acperience

探讨了在资源受限设备上运行复杂卷积神经网络(CNN)的挑战,并提出了一种简化方法,通过将权重二值化来减少计算需求。

题目链接:HDU5734

Acperience

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 746    Accepted Submission(s): 402


Problem Description
Deep neural networks (DNN) have shown significant improvements in several application domains including computer vision and speech recognition. In computer vision, a particular type of DNN, known as Convolutional Neural Networks (CNN), have demonstrated state-of-the-art results in object recognition and detection.

Convolutional neural networks show reliable results on object recognition and detection that are useful in real world applications. Concurrent to the recent progress in recognition, interesting advancements have been happening in virtual reality (VR by Oculus), augmented reality (AR by HoloLens), and smart wearable devices. Putting these two pieces together, we argue that it is the right time to equip smart portable devices with the power of state-of-the-art recognition systems. However, CNN-based recognition systems need large amounts of memory and computational power. While they perform well on expensive, GPU-based machines, they are often unsuitable for smaller devices like cell phones and embedded electronics.

In order to simplify the networks, Professor Zhang tries to introduce simple, efficient, and accurate approximations to CNNs by binarizing the weights. Professor Zhang needs your help.

More specifically, you are given a weighted vector  W=(w1,w2,...,wn) . Professor Zhang would like to find a binary vector  B=(b1,b2,...,bn)   (bi{+1,1}) and a scaling factor  α0  in such a manner that  WαB2  is minimum.

Note that   denotes the Euclidean norm (i.e.  X=x21++x2n , where  X=(x1,x2,...,xn) ).
 

Input
There are multiple test cases. The first line of input contains an integer  T , indicating the number of test cases. For each test case:

The first line contains an integers  n   (1n100000)  -- the length of the vector. The next line contains  n  integers:  w1,w2,...,wn   (10000wi10000) .
 

Output
For each test case, output the minimum value of  WαB2  as an irreducible fraction " p / q " where  p q  are integers,  q>0 .
 

Sample Input
  
3 4 1 2 3 4 4 2 2 2 2 5 5 6 2 3 4
 

Sample Output
  
5/1 0/1 10/1
 

Author
zimpha
 

Source
 

Recommend
wange2014   |   We have carefully selected several similar problems for you:   5746  5743  5741  5740  5739 
 

题意:大概就是求方差,不过是求所有数据绝对值的方差。

题目分析:题目数据量是10000*1000,不过如果普通的求方差的做法求平方再求和的话ull也装不下,因此需要优化公式,这个不难化,最后的公式是(直接copy了下)∑i=1nwi2−1n(∑i=1n∣wi∣)2i=1nwi2n1(i=1nwi)2
 

//
//  main.cpp
//  Acperience
//
//  Created by teddywang on 2016/7/21.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long int T,n;
long long int num[100010];

long long int gcd(long long int a,long long int b)
{
    return a%b==0?b:gcd(b,a%b);
}

int main()
{
    cin>>T;
    while(T--)
    {
        cin>>n;
        long long int ans1=0,ans2=0;
        for(long long int i=0;i<n;i++)
        {
            scanf("%lld",&num[i]);
            if(num[i]<0) num[i]=-num[i];
            ans1+=num[i];
            ans2+=num[i]*num[i];
        }
        long long int buf=ans1*ans1;
        long long int c=gcd(buf,n);
        buf/=c;
        n/=c;
        printf("%lld/%lld\n",ans2*n-buf,n);
    }
}


内容概要:本文详细介绍了“秒杀商城”微服务架构的设计与实战全过程,涵盖系统从需求分析、服务拆分、技术选型到核心功能开发、分布式事务处理、容器化部署及监控链路追踪的完整流程。重点解决了高并发场景下的超卖问题,采用Redis预减库存、消息队列削峰、数据库乐观锁等手段保障数据一致性,并通过Nacos实现服务注册发现与配置管理,利用Seata处理跨服务分布式事务,结合RabbitMQ实现异步下单,提升系统吞吐能力。同时,项目支持Docker Compose快速部署和Kubernetes生产级编排,集成Sleuth+Zipkin链路追踪与Prometheus+Grafana监控体系,构建可观测性强的微服务系统。; 适合人群:具备Java基础和Spring Boot开发经验,熟悉微服务基本概念的中高级研发人员,尤其是希望深入理解高并发系统设计、分布式事务、服务治理等核心技术的开发者;适合工作2-5年、有志于转型微服务或提升架构能力的工程师; 使用场景及目标:①学习如何基于Spring Cloud Alibaba构建完整的微服务项目;②掌握秒杀场景下高并发、超卖控制、异步化、削峰填谷等关键技术方案;③实践分布式事务(Seata)、服务熔断降级、链路追踪、统一配置中心等企业级中间件的应用;④完成从本地开发到容器化部署的全流程落地; 阅读建议:建议按照文档提供的七个阶段循序渐进地动手实践,重点关注秒杀流程设计、服务间通信机制、分布式事务实现和系统性能优化部分,结合代码调试与监控工具深入理解各组件协作原理,真正掌握高并发微服务系统的构建能力。
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
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
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
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