## 10690 - Expression Again

All about problems in Volume 106. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

Moderator: Board moderators

ImLazy
Experienced poster
Posts: 215
Joined: Sat Jul 10, 2004 4:31 pm
Location: Shanghai, China

### 10690

To maximize the expression, I think I should divide the numbers into two parts and make their difference minimum. But how to do it? Some one says DP. But I still have no idea how to DP. Who can give more hints.
I stay home. Don't call me out.

Martin Macko
A great helper
Posts: 481
Joined: Sun Jun 19, 2005 1:18 am
Location: European Union (Slovak Republic)

### Re: How to solve 10690?

There is already a thread on this problem. If there is a thread on a particular problem, please, use it to post your question and do not create a new one. (see http://online-judge.uva.es/board/viewtopic.php?t=6213 and http://online-judge.uva.es/board/viewtopic.php?t=6233)
forum 'Volume CVI' description wrote:All about problems in Volume CVI. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

ImLazy
Experienced poster
Posts: 215
Joined: Sat Jul 10, 2004 4:31 pm
Location: Shanghai, China
OK.
Sorry.
I stay home. Don't call me out.

ImLazy
Experienced poster
Posts: 215
Joined: Sat Jul 10, 2004 4:31 pm
Location: Shanghai, China
How to DP? Could you tell it more detailedly? I've solved 10664, but keep getting TLE on 10690. Let me describe my algorithm:

Code: Select all

``````Use an array of set, named "reach". reach[i] stores the sums that can be reached by i numbers.

for every number n {
for every count of number c (from greater to lower) {
With the set reach[c] and the number n, I can add more sums to the set reach[c + 1].
}
}``````
And I used an optimizing: in the second loop, the count number c is never greater than the count of x in the problem description (Or the count of y, use the lower one).
Can you give me more idea to optimize my algorithm? Or my algorithm is totally wrong?
I stay home. Don't call me out.

ImLazy
Experienced poster
Posts: 215
Joined: Sat Jul 10, 2004 4:31 pm
Location: Shanghai, China
Who can answer me?
I stay home. Don't call me out.

smilitude
Experienced poster
Posts: 137
Joined: Fri Jul 01, 2005 12:21 am
I am not that sure , but did you notice that they asked that the numbers can be negetive ? I think that can produce wrong answers for your algo..

Like if you have -39 to add from greater value, like 1045, you'll have 1006, then again you'll have 967...
fahim
#include <smile.h>

New poster
Posts: 17
Joined: Thu Jul 03, 2014 11:13 am

### Re: 10690 - Expression Again

Hi, I got time limit exceeded for this problem. Please tell me how to improve solution. I used dp subset sum.

Code: Select all

``````#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int _min,_max;
int dp[110][110][5500];
int N,M,a[110];
int mi,totalsum;
void ExpAgain(int idx,int num,int sum){
if(num==mi){
_min=min(_min,sum*(totalsum-sum));
_max=max(_max,sum*(totalsum-sum));
return;
}
if(idx>=(N+M)) return;
if(dp[idx][num][sum+2600]!=-1) return;
dp[idx][num][sum+2600]=1;
ExpAgain(idx+1,num+1,sum+a[idx]);
ExpAgain(idx+1,num,sum);
}
int main()
{
//freopen("in.txt","r",stdin);
while(scanf("%d %d",&N,&M)!=EOF){
totalsum=0;
for(int i=0;i<(N+M);i++){
scanf("%d",&a[i]);
totalsum+=a[i];
}
mi=min(N,M);
_max=0;
_min=2000000000;
memset(dp,-1,sizeof dp);
ExpAgain(0,0,0);
printf("%d %d\n",_max,_min);
}
}
``````

brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA

### Re: 10690 - Expression Again

My approach, AC in 1.156 seconds.

You need to divide the numbers into two groups, so find all the possible sums of min(N, M) of the integers. The sum of the remaining integers is easy to find.
I use a hash: unordered_set<int> to store all possible sums of each number of integers.
Iterate through each of the N + M integers, and then iterate through each of the numbers of integers, and add the current integer to each of the possible sums.
Check input and AC output for thousands of problems on uDebug!

metaphysis
Experienced poster
Posts: 139
Joined: Wed May 18, 2011 3:04 pm

### Re: 10690 - Expression Again

Test data generator.

Code: Select all

``````#include <iostream>
#include <cstdlib>
#include <ctime>

using namespace std;

int main(int argc, char *argv[])
{
srand(time(NULL));
int number[101];

for (int i = 1; i <= 110; i++)
{
int N = rand() % 50; N++;
int M = rand() % 50; M++;
int T = (N + M);

cout << N << ' ' << M << '\n';

for (int j = 1; j <= T; j++)
{
int K = rand() % 101; K -= 50;
number[j] = K;
}

cout << number[1];
for (int j = 2; j <= T; j++)
cout << ' ' << number[j];
cout << '\n';
}

return 0;
}
``````