hdu-5734 Acperience(数学)

最新推荐文章于 2019-01-03 22:30:11 发布

转载最新推荐文章于 2019-01-03 22:30:11 发布 · 90 阅读

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原文链接：http://www.cnblogs.com/zhangchengc919/p/5692885.html

本文探讨了在资源受限的设备上运行复杂卷积神经网络（CNN）的问题，并提出了一种通过权重二值化来简化网络的方法。该方法旨在寻找一个二进制向量和缩放因子以最小化原始权重与二值化权重之间的欧式距离。

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题目链接：

Acperience

Time Limit: 4000/2000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

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

题意：

问∑(w[i]+x[i]*a)^2最小是多少？x[i]=1或-1;a>0;

思路：

∑(w[i]+x[i]*a)^2=∑w[i]^2+n*a^2+2*a*∑x[i]w[i];

∑w[i]^2是常数,现在变成了一个一元二次函数,找出它的最小值;

根据二次函数的的性质,∑x[i]w[i]尽量大,所以就很简单了;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-6;

int a[N],n;
LL sum=0,ans=0;
 LL gcd(LL x,LL y)
 {
     if(y==0)return x;
     return gcd(y,x%y);
 }
int main()
{
        int t;
        read(t);
        while(t--)
        {
            read(n);
            sum=0,ans=0;
            For(i,1,n)
            {
                read(a[i]);
                sum=sum+(LL)a[i]*a[i];
                if(a[i]<0)ans=ans-a[i];
                else ans=ans+a[i];
            }
            LL x=(LL)n;
            LL g=gcd(x*sum-ans*ans,x);
            cout<<(x*sum-ans*ans)/g<<"/"<<x/g<<endl;


        }
        return 0;
}

转载于:https://www.cnblogs.com/zhangchengc919/p/5692885.html