hdu5734 Acperience（数学）

Acperience

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 516    Accepted Submission(s): 278

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 integerT, 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

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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我的：

  这个题目最重要的就是：，然后这个式子看成二元一次方程组，再注意一点细节就好了。

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<algorithm>#include<stdlib.h>#include<queue>#include<stack>#include<set>#include<vector>#include<map>using namespace std;const int maxn=100000+10;typedef long long ll;//int gcd(int x,int y)//{//    if(y==0)//        return x;//    return gcd(y,x%y);//}ll gcd(ll x,ll y){    if(y==0)        return x;    return gcd(y,x%y);}ll work(ll x){    if(x<0)        return -x;    return x;}int main(){    int t;    cin>>t;    while(t--)    {        int n;        cin>>n;        ll w[maxn];        ll  absolute_value=0,quare_sum=0;        for(int i=0;i<n;i++)            cin>>w[i],absolute_value+=work(w[i]),quare_sum+=w[i]*w[i];        ll a=absolute_value*absolute_value;        ll temp=gcd(a,n);        a=a/temp;        n=n/temp;        a=quare_sum*n-a;        temp=gcd(n,a);        a=a/temp;        n=n/temp;        cout<<a<<"/"<<n<<endl;    }}

美素数hdu4548

水题，直贴代码：

#include<bits/stdc++.h>using namespace std;const int maxn = 1000000 + 10;bool is_prime[maxn];int prime[maxn];bool vis[maxn];int a[maxn];void Init(){    memset(is_prime,true,sizeof(is_prime));    is_prime[0] = is_prime[1] = false;    for(int i = 2; i <maxn ; ++ i)    {        if(is_prime[i])            for(int j = i * 2; j < maxn ; j += i)        {            is_prime[j] = false;        }    }    memset(vis,false,sizeof(vis));    for(int i = 1; i < maxn ; ++ i)    {        if(is_prime[i])        {            int sum = 0;            int t = i;            while(t)            {                sum += t % 10;                t /= 10;            }            if(is_prime[sum])            {                vis[i] = true;            }        }    }    a[0] = 0;    for(int i = 1; i < maxn ; ++ i)    {        if(vis[i])        {            a[i] = a[i - 1] + 1;        }        else a[i] = a[i - 1];    }}int main(){    Init();    int Tcase;    scanf("%d",&Tcase);    for(int ii = 1; ii <= Tcase ; ++ ii)    {//        for(int i = 1; i < 100 ; ++ i)//        cout << a[i] << " ";//        cout << endl;        int x,y;        scanf("%d%d",&x,&y);//        cout << a[y]<< " "<<a[x - 1] << endl;        printf("Case #%d: %d\n",ii,a[y] - a[x - 1]);    }    return 0;}