2016 Multi-University Training Contest 2

Acperience

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

Total Submission(s): 748 Accepted Submission(s): 404



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

一元二次不等式最值公式

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
const int maxn=100010;
typedef long long LL;
LL num[maxn];
LL gcd(LL a,LL b){
return b==0?a:gcd(b,a%b);
}
int main()
{
int t,n;cin>>t;
while(t--){
scanf("%d",&n);
LL sum=0,qsum=0;
for(int i=1;i<=n;++i){
scanf("%lld",&num[i]);
sum+=abs(num[i]);qsum+=num[i]*num[i];
}
//LL ans = (4*a*c-b*b)/4*a;
LL u=(4*qsum*n-4*sum*sum);
LL d=4*n;
LL GCD=gcd(u,d);
if(u==0){
printf("0/1\n");
}
else {
printf("%lld/%lld\n",u/GCD,d/GCD);
}

}
return 0;
}

It's All In The Mind

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 855 Accepted Submission(s): 399



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

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
int num[110];
int gcd(int a,int b){
return b==0?a:gcd(b,a%b);
}
int main()
{
int t,n,m;cin>>t;
while(t--){
scanf("%d%d",&n,&m);
memset(num,-1,sizeof(num));
int x,y;
for(int i=1;i<=m;++i){
scanf("%d%d",&x,&y);
num[x]=y;
}
if(num[1]!=-1){
if(num[2]==-1){
num[2]=num[1];
}
}
else {
num[1]=100;
if(num[2]==-1)num[2]=100;
}
int k=0;
for(int i=n;i>=3;--i){
if(num[i]==-1)num[i]=k;
else k=num[i];
}
int u=num[1]+num[2],d=0;
for(int i=1;i<=n;++i){
d+=num[i];
}
int GCD=gcd(u,d);
printf("%d/%d\n",u/GCD,d/GCD);
}
return 0;
}

Keep On Movin

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

Total Submission(s): 533 Accepted Submission(s): 386



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

题意：给出N种不同字符及每种字符的数量求能够组成的回文字符串的长度的最小值最大化

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
const int maxn=100010;
int num[maxn];
int main()
{
int t,n;cin>>t;
while(t--){
scanf("%d",&n);
int odd=0,sum=0;
for(int i=1;i<=n;++i){
scanf("%d",&num[i]);
sum+=num[i];
if(num[i]&1)odd++;
}
if(odd==0){
printf("%d\n",sum);
}
else {
int even=(sum-odd)/2;
printf("%d\n",(even/odd)*2+1);
}
}
return 0;
}

La Vie en rose

Time Limit: 14000/7000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 878 Accepted Submission(s): 474



Problem Description

Professor Zhang would like to solve the multiple pattern matching problem, but he only has only one pattern string p=p1p2...pm.
So, he wants to generate as many as possible pattern strings from p using
the following method:

1. select some indices i1,i2,...,ik such
that 1≤i1<i2<...<ik<|p| and |ij−ij+1|>1 for
all 1≤j<k.

2. swap pij and pij+1 for
all 1≤j≤k.

Now, for a given a string s=s1s2...sn,
Professor Zhang wants to find all occurrences of all the generated patterns in s.

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 two integers n and m (1≤n≤105,1≤m≤min{5000,n}) --
the length of s and p.

The second line contains the string s and
the third line contains the string p.
Both the strings consist of only lowercase English letters.

Output

For each test case, output a binary string of length n.
The i-th
character is "1" if and only if the substring sisi+1...si+m−1 is
one of the generated patterns.

Sample Input

3
4 1
abac
a
4 2
aaaa
aa
9 3
abcbacacb
abc

Sample Output

1010
1110
100100100

Author

zimpha

Source

2016 Multi-University Training Contest 2

暴力即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#define PI acos(-1.0)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
char str[100010];
char p[5100];
int n,m;
int main() {
int t;cin>>t;
while(t--)
{
scanf("%d%d",&n,&m);
scanf("%s%s",str,p);
for(int i=0;str[i];++i){
bool flag=true;int pos=i;
for(int j=0;p[j];++j){
if(str[pos] != p[j]){
if(str[pos+1] == p[j] && p[j+1] == str[pos]){
j++; pos++;
}
else {
flag=false;break;
}
}
pos++;
}
printf("%d",flag);
}
printf("\n");
}
return 0;
}