Acperience HDU - 5734

本文探讨了在深度学习领域中,针对卷积神经网络(CNN)的权重进行二值化处理的方法及其应用。该研究旨在减少CNN模型的内存占用和计算需求,使之更适用于移动设备等资源受限的场景。

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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)W=(w1,w2,...,wn). Professor Zhang would like to find a binary vector B=(b1,b2,...,bn)B=(b1,b2,...,bn) (bi{+1,1})(bi∈{+1,−1}) and a scaling factor α0α≥0in such a manner that WαB2‖W−αB‖2 is minimum. 

Note that ‖⋅‖ denotes the Euclidean norm (i.e. X=x21++x2n−−−−−−−−−−−√‖X‖=x12+⋯+xn2, where X=(x1,x2,...,xn)X=(x1,x2,...,xn)).

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

The first line contains an integers n(1n100000)(1≤n≤100000) -- the length of the vector. The next line contains nn integers: w1,w2,...,wnw1,w2,...,wn (10000wi10000)(−10000≤wi≤10000).
OutputFor each test case, output the minimum value of WαB2‖W−αB‖2 as an irreducible fraction "pp/qq" where pp, qq are integers, q>0q>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

一个公式推导题
最后的是式子为n* (x1^2+x2^2+…….+xn^2) - sum*sum
 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include <vector>
 5 #include<math.h>
 6 using namespace std;
 7 long long gcd(long long  a,long long b)
 8 {
 9     if (a==0) return b;
10     else gcd(b%a,a);
11 }
12 int main()
13 {
14     int t;
15     while(scanf("%d",&t)!=EOF){
16         while(t--){
17             int n,a;
18             scanf("%d",&n);
19             long long ans=0,cnt=0,sum=0;
20             for (int i=0 ;i<n ;i++){
21                 scanf("%d",&a);
22                 a=abs(a);
23                 ans+=a;
24                 cnt+=a*a;
25             }
26             ans=ans*ans;
27             sum=n*cnt-ans;
28             long long b=gcd(sum,n);
29             printf("%lld/%lld\n",sum/b,n/b);
30         }
31     }
32     return 0;
33 }

 

 

转载于:https://www.cnblogs.com/qldabiaoge/p/8512088.html

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