hdoj 5734 5742 5744 2016多校2（复现）<数学---思维----贪心>

Acperience

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 972    Accepted Submission(s): 517


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

化为w1 ^2+w2 ^2+w3 ^2+...+wn ^2+n b^2  -   2sb

b在s/n处使函数最小------

记得分子分母及时化简-----不然爆long long

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
LL gcd(LL dd,LL xx)
{
LL c;
if (dd<xx)
{
c=dd;dd=xx;xx=c;
}
return dd%xx==0? xx:gcd(xx,dd%xx);
}
LL shu[1212120];
void s()
{
LL s=0,n;
scanf("%lld",&n);
for (int i=0;i<n;i++)
{
scanf("%lld",&shu[i]);
if (shu[i]<0)
shu[i]=0-shu[i];
s+=shu[i];
}
sort(shu,shu+n);
if (shu[0]==shu[n-1])
{
printf("0/1\n");
return ;
}
LL ss=0,mu,zhi;
for (int i=0;i<n;i++)
ss+=shu[i]*shu[i];
zhi=s*s;mu=n;
// printf("%lld  %lld      %lld\n",zhi,mu,ss);
LL lp=gcd(zhi,mu);
zhi/=lp;mu/=lp;
zhi=mu*ss-zhi;
lp=gcd(zhi,mu);
zhi/=lp;mu/=lp;
printf("%lld/%lld\n",zhi,mu);
}
int main()
{
int n;scanf("%d",&n);
while (n--)
s();
return 0;
}

多校联合训练的同学们~

It's All In The Mind Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1119    Accepted Submission(s): 546 Problem Description Professor Zhang has a number sequence a1,a2,...,an. However, the sequence is not complete and some elements are missing. Fortunately, Professor Zhang remembers some properties of the sequence: 1. For every i∈{1,2,...,n}, 0≤ai≤100. 2. The sequence is non-increasing, i.e. a1≥a2≥...≥an. 3. The sum of all elements in the sequence is not zero. Professor Zhang wants to know the maximum value of a1+a2∑ni=1ai among all the possible sequences.   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 contains two integers n and m (2≤n≤100,0≤m≤n) -- the length of the sequence and the number of known elements. In the next m lines, each contains two integers xi and yi (1≤xi≤n,0≤yi≤100,xi<xi+1,yi≥yi+1), indicating that axi=yi.   Output For each test case, output the answer as an irreducible fraction "p/q", where p, q are integers, q>0.   Sample Input 2 2 0 3 1 3 1   Sample Output 1/1 200/201   Author zimpha   Source 2016 Multi-University Training Contest 2

让a1+a2尽可能的大-----

s尽可能的小

代码：

#include<cstdio>
#include<cstring>
int gcd(int dd,int xx)
{
return dd%xx==0? xx:gcd(xx,dd%xx);
}
void s()
{
int n,m,a,b;int fafe[120];int shu[120];
memset(fafe,true,sizeof(fafe));
scanf("%d%d",&n,&m);
while (m--)
{
scanf("%d%d",&a,&b);
shu[a]=b;
fafe[a]=false;
}
if (fafe[1]&&fafe[2])
{
shu[1]=shu[2]=100;
}
else if (fafe[1])
{
shu[1]=100;
}
else if (fafe[2])
shu[2]=shu[1];
int mi=0;
for (int i=n;i>2;i--)
{
if (!fafe[i])
mi=shu[i];
else
shu[i]=mi;
}
int zhi=shu[1]+shu[2];
int mu=0;
for (int i=1;i<=n;i++)
mu+=shu[i];
if (zhi==0)
{
printf("0/1\n");
return ;
}
int lp=gcd(mu,zhi);
mu/=lp;zhi/=lp;
printf("%d/%d\n",zhi,mu);
}
int main()
{
int n;scanf("%d",&n);
while (n--)
s();
return 0;
}

Keep On Movin

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 726    Accepted Submission(s): 526


Problem Description

Professor Zhang has kinds of characters and the quantity of the i-th
character is ai.
Professor Zhang wants to use all the characters build several palindromic strings. He also wants to maximize the length of the shortest palindromic string.

For example, there are 4 kinds of characters denoted as 'a', 'b', 'c', 'd' and the quantity of each character is {2,3,2,2} .
Professor Zhang can build {"acdbbbdca"}, {"abbba", "cddc"}, {"aca", "bbb", "dcd"}, or {"acdbdca", "bb"}. The first is the optimal solution where the length of the shortest palindromic string is 9.

Note that a string is called palindromic if it can be read the same way in either direction.

 

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 integer n (1≤n≤105) --
the number of kinds of characters. The second line contains n integers a1,a2,...,an (0≤ai≤104).

 

Output

For each test case, output an integer denoting the answer.

 

Sample Input

4
4
1 1 2 4
3
2 2 2
5
1 1 1 1 1
5
1 1 2 2 3

 

Sample Output

3
6
1
3

 

Author

zimpha

 

Source

2016 Multi-University Training Contest 2

看字符奇数的个数====然后让偶数尽可能的平均补奇数---

代码“：

#include<cstdio>
void s()
{
int n,c;scanf("%d",&n);
int a=0,b=0;
while (n--)
{
scanf("%d",&c);
a+=c%2;
b+=c/2;
}
if (a)
c=1+b/a*2;
else
c=b*2;
printf("%d\n",c);
}
int main()
{
int n;scanf("%d",&n);
while (n--)
s();
return 0;
}