hdu 5734 Acperience（2016多校第二场）

最新推荐文章于 2025-08-11 22:14:14 发布

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原文链接：http://www.cnblogs.com/pshw/p/5694520.html

文章标签：

#java #人工智能

本文探讨了在资源受限设备上应用卷积神经网络(CNN)的问题，通过权重二值化来简化网络结构并减少计算资源的需求。文章提供了一种寻找最佳二值权重和缩放因子的方法，并附带实现该优化方案的源代码。

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Acperience

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 484 Accepted Submission(s): 258

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

Input

There are multiple test cases. The first line of input contains an integer

Output

For each test case, output the minimum value of

Sample Input

1 2 3 4

2 2 2 2

5 6 2 3 4

Sample Output

5/1

0/1

10/1

Author

zimpha

题意：

给定w1-wn的值，α>=0，B可以等于1或者-1，求这个的最小值。定义运算：

。

数学推论题，可以推出数学计算公式。

（w1-aB1）^2+（w2-aB2）^2+...（wn-aBn）^2

w1的取值范围是−10000≤wi≤10000，由于B1既可以取1也可以取-1，因此可以将所有的负数变为正数考虑。

因此：（|w1|-a）^2+（|w2|-a）^2+...（|wn|-a）^2（以后的w 全部为|w|，麻烦不想写）

又因为要取最小值，显然 a=（w1+w2+...+wn）/n

展开式子： w1^2+a^2-2aw1

　　　　　 w2^2+a^2-2aw2

......

wn^2+a^2-2awn

得：（w1^2+w2^2+...+wn^2）+na^2-2a(w1+w2+...+wn)

将w1^2+w2^2+...+wn^2看作s1，w1+w2+...+wn看作s2，则：s1+na^2-2as2

且a=（w1+w2+...+wn）/n=s2/n

得： s1+n(s2/n)^2-2(s2/n)*s2=s1+s2^2/n-2(s2^2)/n=(ns1-s2^2)/n

附上代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#define  ll long long
using namespace std;

ll _gcd(ll x,ll y)
{
    ll z;
    if(x<y) z=x,x=y,y=z;
    while(y)
    {
        z=x%y;
        x=y;
        y=z;
    }
    return x;
}

int xabs(int x)
{ 
    return x>0?x:-x;
}
int main()
{
    int T,i,j,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ll aa,bb,t;
        ll s1=0,s2=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&m);
            m=xabs(m);
            s1+=m*m;
            s2+=m;
        }
        aa=n*s1-s2*s2;
        bb=n;
        t=_gcd(aa,bb);
        printf("%lld/%lld\n",aa/t,bb/t);
    }
    return 0;
}