hdu 5734 Acperience

Acperience

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 617 Accepted Submission(s): 329

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

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

2016 Multi-University Training Contest 2

Acperience

展开式子, $ight|^2=displaystylealpha^2sum_{i=1}^{n}b_i^2-2alphasum_{i=1}^n{w_ib_i}+sum_{i=1}^{n}w_i^2$ .

由于 $b_iin {+1,-1}$ , 那么 $displaystylesum_{i=1}^{n}b_i^2=n$ , 显然 $c=displaystylesum_{i=1}^{n}w_i^2$ 也是常数. 转化成求 $displaystylealpha^2n-2alphasum_{i=1}^n{w_ib_i}+c$ 的最小值. 对于固定的 $α > 0$ , 只要 $displaystylesum_{i=1}^n{w_ib_i}$ 最大就好了. 显然 $b_i=sign(w_i)$ 的时候, $displaystylesum_{i=1}^n{w_ib_i}=sum_{i=1}^{n}|w_i|$ 最大. 进一步的, 上面显然是一个关于 $α$ 的二次方程, 于是当 $alpha=frac{1}{n}displaystylesum_{i=1}^n{w_ib_i}=frac{1}{n}displaystylesum_{i=1}^{n}{|w_i|}$ 时, 取到最大值.

化简下, 可以得到最小值是 $sum_{i=1}^n{w_i^2}-frac{1}{n}(displaystylesum_{i=1}^{n}|w_i|)^2$

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<cmath>
using namespace std;
unsigned long long gcd(unsigned long long x,unsigned long long y)
{
    unsigned long long a;
    if(x<y) a=x,x=y,y=a;
    while(y!=0){
        a=x%y;
        x=y;
        y=a;
    }
    return x;
    
}
int main(){
    unsigned long long sum1=0;
    unsigned long long sum2=0;
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        sum1=0;sum2=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            int tmp;
            scanf("%d",&tmp);
            sum1+=(tmp*tmp);
            sum2+=abs(tmp);
        }
        sum2*=sum2;
        sum1*=n;
        unsigned long long div=(gcd( sum1-sum2,n));
        printf("%lld/%lld
",(sum1-sum2)/div,n/div);
    }
    return 0;
}

$ight|^2=displaystylealpha^2sum_{i=1}^{n}b_i^2-2alphasum_{i=1}^n{w_ib_i}+sum_{i=1}^{n}w_i^2$