HDU5734 Acperience
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题目链接:HDU5734
Acperience
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 746 Accepted Submission(s): 402
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 vectorW=(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) ).
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
Note that
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 integersn (1≤n≤100000) -- the length of the vector. The next line contains n integers: w1,w2,...,wn (−10000≤wi≤10000) .
The first line contains an integers
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
341 2 3 442 2 2 255 6 2 3 4
Sample Output
5/10/110/1
Author
zimpha
Source
2016 Multi-University Training Contest 2
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wange2014 | We have carefully selected several similar problems for you: 5746 5743 5741 5740 5739
题意:大概就是求方差,不过是求所有数据绝对值的方差。
题目分析:题目数据量是10000*1000,不过如果普通的求方差的做法求平方再求和的话ull也装不下,因此需要优化公式,这个不难化,最后的公式是(直接copy了下)∑i=1nwi2−n1(i=1∑n∣wi∣)2
//// main.cpp// Acperience//// Created by teddywang on 2016/7/21.// Copyright © 2016年 teddywang. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;long long int T,n;long long int num[100010];long long int gcd(long long int a,long long int b){ return a%b==0?b:gcd(b,a%b);}int main(){ cin>>T; while(T--) { cin>>n; long long int ans1=0,ans2=0; for(long long int i=0;i<n;i++) { scanf("%lld",&num[i]); if(num[i]<0) num[i]=-num[i]; ans1+=num[i]; ans2+=num[i]*num[i]; } long long int buf=ans1*ans1; long long int c=gcd(buf,n); buf/=c; n/=c; printf("%lld/%lld\n",ans2*n-buf,n); }}
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