【HDU5734 2016 Multi-University Training Contest 2A】【公式代入推導】Acperience n維向量各有加減最小模長

snowy_smile發表於2016-07-28
AI公式

Acperience

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 993    Accepted Submission(s): 532


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




 


Author


zimpha


 


Source


2016 Multi-University Training
Contest 2




#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 100010, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int casenum, casei;
LL n;
LL a[N];
LL gcd(LL x, LL y)
{
	return y == 0 ? x : gcd(y, x%y);
}
/*
n * ∑(wi^2) - (∑|wi|)^2 (1e18-1e16)
-----------------------------------
                n
*/
int main()
{
	scanf("%d", &casenum);
	for (casei = 1; casei <= casenum; ++casei)
	{	
		LL a = 0;
		LL b = 0;
		scanf("%d", &n);
		for (int i = 1; i <= n; ++i)
		{
			LL x; scanf("%lld", &x);
			a += x*x;
			b += abs(x);
		}
		LL top = n*a - b*b;
		LL bot = n;
		LL g = gcd(top, bot);
		printf("%lld/%lld\n", top/g, bot/g);
	}
	return 0;
}
/*
【trick&&吐槽】
1，很多題目看起來很冗雜，
其實我們讀重點的話，就會發現題目是可以入手的。
比賽的時候也不能放棄題目閱讀，往往藏了很多簡單題。

2，一定要注意資料上限，隊友推出了一個爆LL的錯誤公式

3，java做大數運算是很耗時間的，儘量規避。

【題意】
給你一個n維向量a，
然後我們希望對a向量的每個維度都加減一個相同的值，
使得得到的目標向量b的模儘可能小。

【型別】
公式化簡

【分析】
如何做公式化簡？
先一股腦地做代入！
(w1-ab1, w2-ab2, ... , wn-ab3)，然後得到——
原式=
 (w1-ab1)^2
+(w2-ab2)^2
+(w3-ab3)^2
...
+(wn-abn)^2
=(w1^2+...+wn^2)[定值] + (a^2)*(b1^2+b2^2+...+bn^2) - 2a(w1b1+w2b2+...+wnbn)
=[定值]+(a^2)*n-(2a)*(w1b1+w2b2+...+wnbn)
我們希望使其儘可能小，
顯然 要使得（w1b1+w2b2+...+wnbn）儘可能大，這個很容易做的。使其變為了給定的定值。

然後，我們最後需要考慮的是一個
(x^2)*n-2x*[定值]+[定值]的一元二次方程。
顯然，其最小值為
    b^2                (∑|wi|)^2           n * ∑(wi^2) - (∑|wi|)^2 (1e18-1e16)
c- ----  = ∑(wi^2) -  ----------      =  ----------------------------
    4a                   n                              n

【時間複雜度&&優化】
O（n）

*/


 




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