HDU 5734 Acperience(数学推导)
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.
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 (1≤n≤100000) -- the length of the vector. The next line contains n integers: w1,w2,...,wn (−10000≤wi≤10000).
Output
3
4
1 2 3 4
4
2 2 2 2
5
5 6 2 3 4
5/1
0/1
10/1
题意
给你一个n维向量w,求∥W−αB∥2的最小值,其中B=(b1,b2,...,bn) (bi∈{+1,−1}),α≥0
题解
开始误以为是平均数最小,WA了几次后开始推式子
min(∥w−αb∥2)=min(∑(wi2-2αbiwi+α2bi2))
由于bi∈{+1,−1},易得bi*w≥0
=min(∑(wi2-2α|wi|+α2))=min(∑(α2-2α|wi|+wi2))=min(nα2-2α∑|wi|+∑wi2)
可知当α=∑|wi|/n时函数取到min
代入化简得=-(∑|wi|)2/n+∑wi2
通分=(n∑wi2-(∑|wi|)2)/n
gc=gcd(n∑wi2-(∑|wi|)2,n)
所以p=(n∑wi2-(∑|wi|)2)/gc,q=n/gc
代码
#include<bits/stdc++.h>
using namespace std; #define ll long long
const int maxn=1e5+;
int a[maxn];
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
ll sum=,ac=;
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=abs(a[i]);
ac+=a[i]*1LL*a[i];
}
ll gc=__gcd(ac*n-sum*sum,1LL*n);
printf("%lld/%lld\n",(ac*n-sum*sum)/gc,n/gc);
}
return ;
}
HDU 5734 Acperience(数学推导)的更多相关文章
- HDU 5734 Acperience (推导)
Acperience 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5734 Description Deep neural networks (DN ...
- HDU 5734 Acperience ( 数学公式推导、一元二次方程 )
题目链接 题意 : 给出 n 维向量 W.要你构造一个 n 维向量 B = ( b1.b2.b3 ..... ) ( bi ∈ { +1, -1 } ) .然后求出对于一个常数 α > 0 使得 ...
- HDU 5734 Acperience(返虚入浑)
p.MsoNormal { margin: 0pt; margin-bottom: .0001pt; text-align: justify; font-family: Calibri; font-s ...
- hdu 5734 Acperience 水题
Acperience 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5734 Description Deep neural networks (DN ...
- HDU 5734 Acperience
Acperience Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total ...
- hdu 5734 Acperience(2016多校第二场)
Acperience Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total ...
- hdu.5211.Mutiple(数学推导 && 在logn的时间内求一个数的所有因子)
Mutiple Accepts: 476 Submissions: 1025 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 6553 ...
- HDU 5734 Acperience (公式推导) 2016杭电多校联合第二场
题目:传送门. #include <iostream> #include <algorithm> #include <cstdio> #include <cs ...
- HDU 5984 题解 数学推导 期望
Let’s talking about something of eating a pocky. Here is a Decorer Pocky, with colorful decorative s ...
随机推荐
- python------Socket网略编程
上节课程回顾: 静态变量:与类无关,不能访问类里的任何属性和方法. 类方法:只能访问类变量. 属性:把一个方法变成静态属性, 反射: __new__:先于__init__执行: __call__: c ...
- 开发一个简单的postgresql extension
主要是学习如何编写一个简单的pg extension,参考https://severalnines.com/blog/creating-new-modules-using-postgresql-c ...
- Android related
The build env. ensure that your computer’s BIOS is set up to support Intel’s virtualization extensio ...
- 服务跟踪sleuth和可视化跟踪工具Zipkin
一.增加配置 在Order工程中添加配置 <dependency> <groupId>org.springframework.cloud</groupId> < ...
- CentOS 7下给nginx安装SSL证书
0. DNS要能解析你的网址(域名解析和主机解析,例如example.com和www.example.com都要能解析.注意泛解析记录*.example.com可以存在但在本文中暂时无法用于https ...
- zabbix监控windows服务器
Zabbix agent 在windows上安装部署 1. 下载与解压 地址: http://www.zabbix.com/downloads/2.4.0/zabbix_agents_2.4.0.w ...
- .net core 中间件实战
1.新建一个ASP.NET Core Web Application项目,选择空模板. 2.新建一个类RequestIPMiddleware.cs using Microsoft.AspNetCore ...
- KMS服务器激活WIN方法
KMS激活的过程简单说就是:欲激活的电脑向KMS服务器请求,KMS服务器做出回应同意激活. KMS激活软件是将KMS服务器用一段代码来模拟,做成一个可执行的程序(即所谓的KMS伺服器).KMS激活软件 ...
- Using a ScrollView - RN4
使用滚动条. 1. import import {ScrollView} from "react-native"; 2. Using <ScrollView> ... ...
- 深入了解scanf()/getchar()和gets()/cin等函数
转:http://www.cnblogs.com/FCWORLD/archive/2010/12/04/1896511.html 转:问题描述一:(分析scanf()和getchar()读取字符) s ...