化简-HDU-5734-Acperience

Acperience

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 236 Accepted Submission(s): 125

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

Note that ∥⋅∥ denotes the Euclidean norm (i.e. ∥X∥=sqrt(x1^2+…xn^2), where X=(x1,x2,…,xn)).

Input
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
For each test case, output the minimum value of ∥W−αB∥^2 as an irreducible fraction “p/q” where p, q are integers, q>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

Author
zimpha

Source
2016 Multi-University Training Contest 2

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题意：
噼里啪啦的一堆管理废话，总之就是给定一个数组W，求∥W−αB∥^2的最小值，其中B是与W等大的、元素只为1或-1的数组，α是任意实数。

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题解：
先把题中式子展开来看，就是一个(x1−αB)^2+…(xn−αB)^2，因为B中元素是不定的，那么再进一步展开，为了让原式最小，成了(x1^2+…xn^2)−2α(abs(x1)+…abs(xn))+nα^2，(x1^2+…xn^2)是定值，那么就是求−2α(x1+…xn)+nα^2这部分最小，让s=(abs(x1)+…abs(xn))，这部分合并为nα^2-2sα，那么很明显n大于零，取得最小值点s/n，所以α的值就是s/n。将这个值带入原式，成了(x1^2+…xn^2)-s*s/n。gcd一下后输出就行了。

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//
//  main.cpp
//  HDU-5734-Acperience
//
//  Created by 袁子涵 on 16/7/21.
//  Copyright © 2016年 袁子涵. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>

using namespace::std;
int t;
long long int n,s,fa,tmp,fz,fm,yue;
long long int gcd(long long int a,long long int b)
{
    if (b==0)
        return a;
    return gcd(b, a%b);
}
int main(int argc, const char * argv[]) {
    ios::sync_with_stdio(false);
    cin >> t;
    while (t--) {
        cin >> n;
        s=fa=0;
        for (long long int i=0; i<n ; i++) {
            cin >> tmp;
            fa+=tmp*tmp;
            s+=abs(tmp);
        }
        fz=-s*s+n*fa;
        fm=n;
        yue=gcd(fz, fm);
        cout << fz/yue << '/' << fm/yue << endl;
    }
    return 0;
}