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.
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
3
4
1 2 3 4
4
2 2 2 2
5
5 6 2 3 4
5/1
0/1
10/1
题意
给你一个n维向量w，求∥W−αB∥2的最小值，其中B=(b1,b2,...,bn) (bi∈{+1,−1})，α≥0
题解
开始误以为是平均数最小，WA了几次后开始推式子
min(∥w−αb∥2)=min(∑(wi2-2αbiwi+α2bi2))
由于bi∈{+1,−1}，易得bi*w≥0
=min(∑(wi2-2α|wi|+α2))=min(∑(α2-2α|wi|+wi2))=min(nα2-2α∑|wi|+∑wi2)
可知当α=∑|wi|/n时函数取到min
代入化简得=-(∑|wi|)2/n+∑wi2
通分=(n∑wi2-(∑|wi|)2)/n
gc=gcd(n∑wi2-(∑|wi|)2,n)
所以p=(n∑wi2-(∑|wi|)2)/gc,q=n/gc
代码
#include<bits/stdc++.h>
using namespace std; #define ll long long
const int maxn=1e5+;
int a[maxn];
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
ll sum=,ac=;
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=abs(a[i]);
ac+=a[i]*1LL*a[i];
}
ll gc=__gcd(ac*n-sum*sum,1LL*n);
printf("%lld/%lld\n",(ac*n-sum*sum)/gc,n/gc);
}
return ;
}