hdu 5734 Acperience （数学）

Acperience
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 747 Accepted Submission(s): 403

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∥=x21+⋯+x2n−−−−−−−−−−√, 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

题意：给你$n$个$w$，$n$个$B$，$B$只能取-1或1，让你找一个$α$，使得$\sum_{i=1}^n ||w_i-αB_i||^2$取最小值.

思路：首先，我们来化简题中得到式子，化简得：

$$\sum_{i=1}^n ||w_i-αB_i||^2=\sum_{i=1}^nw_i^2+\sum_{i=1}^nα^2B_i^2+\sum_{i=1}^n-2αB_i$$
设$sum$为$w$的和，$ksum$为$w$的平方和，$α$为$x$，则方程为：$nx^2-2sumx+ksum$，这样就转换成了一个开口向上的一元二次方程式求其最小值的问题，直接求就好了，注意求$sum$的时候要取$|w|$的和，因为$B$可以取-1或者1，所以说$B$可根据的正负调整正负。最后上下化简就可以了。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
ll sum,ksum;
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        int n;scanf("%d",&n);
        sum=0;ksum=0;
        for(int i=1;i<=n;i++)
        {
            int a;scanf("%d",&a);
            sum+=abs(a);
            ksum+=a*a;
        }
        ll a=n,b=-2*sum,c=ksum;
        ll up=n*ksum-sum*sum,down=n;
        ll k=gcd(up,down);
        printf("%I64d/%I64d\n",up/k,down/k);
    }
    return 0;
}