hdu 5734 Acperience（2016多校第二场）

Acperience

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 484    Accepted Submission(s):
258

 

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

 

 

题意：给定w1-wn的值，α>=0，B可以等于1或者-1，求这个的最小值。定义运算：。

 

数学推论题，可以推出数学计算公式。

（w1-aB1）^2+（w2-aB2）^2+...（wn-aBn）^2

w1的取值范围是−10000≤wi≤10000，由于B1既可以取1也可以取-1，因此可以将所有的负数变为正数考虑。

因此：（|w1|-a）^2+（|w2|-a）^2+...（|wn|-a）^2（以后的w 全部为|w|，麻烦不想写）

又因为 要取最小值，显然 a=（w1+w2+...+wn）/n

展开式子 ： w1^2+a^2-2aw1

　　　　　  w2^2+a^2-2aw2

......

wn^2+a^2-2awn

得：（w1^2+w2^2+...+wn^2）+na^2-2a(w1+w2+...+wn)

将w1^2+w2^2+...+wn^2看作s1，w1+w2+...+wn看作s2，则：s1+na^2-2as2

且a=（w1+w2+...+wn）/n=s2/n

得 ： s1+n(s2/n)^2-2(s2/n)*s2=s1+s2^2/n-2(s2^2)/n=(ns1-s2^2)/n

附上代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#define  ll long long
using namespace std;

ll _gcd(ll x,ll y)
{
    ll z;
    if(x<y) z=x,x=y,y=z;
    while(y)
    {
        z=x%y;
        x=y;
        y=z;
    }
    return x;
}

int xabs(int x)
{
    return x>?x:-x;
}
int main()
{
    int T,i,j,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ll aa,bb,t;
        ll s1=,s2=;
        for(i=;i<n;i++)
        {
            scanf("%d",&m);
            m=xabs(m);
            s1+=m*m;
            s2+=m;
        }
        aa=n*s1-s2*s2;
        bb=n;
        t=_gcd(aa,bb);
        printf("%lld/%lld\n",aa/t,bb/t);
    }
    return ;
}