OJ Board

Posted: **Thu Jan 25, 2007 12:54 pm**

It's seems to me the problem of maximum sum of continuous sub sequences of length m.
I know a linear algorithm when length m is not specified , but don't know how to extend it for
specified length m. Can any one kind enough to give me a linear time algorithm for this problem , or any link describing that algorithm.

Posted: **Sat Feb 17, 2007 11:09 am**

It is even easier when m is fixed.
Let a be the number of popes elected in the year i.
Let f[0]=a[0], and f = f[i-1]+a for i>0, so f is a cumulative sum.
Consider the value of f[i+m-1]-f[i-1] for all i.

As for the algorithm for maximum consecutive sum, it is trivial to solve when all the elements of the array is non-negative.

Posted: **Thu Jun 28, 2007 5:36 pm**

Hello..~
Can anybody tell me what is wrong with my code..?
I'm getting WAs.. but I have no idea..

check is the number of occurences of 'i' and..
sum is the cumulative sum up to check

Code: Select all

code removed

Thanks..

Posted: **Thu Jun 28, 2007 6:01 pm**

Try the cases.

Input:

Code: Select all

11
20
20 numbers given in sample :)
12
20
20 numbers given in sample :)
17
20
20 numbers given in sample :)

Output:

Code: Select all

11 12 21
11 12 21
13 6 21

Hope the weird input set helps.

.

Posted: **Thu Jun 28, 2007 6:26 pm**

Thanks a lot..~
I got AC..

PS. very funny input..

It took some time to get the point..

Posted: **Sat Sep 18, 2010 9:29 pm**

I can't realize why I'm getting WA.
Please give me some input output.

Code: Select all

#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
    long int y,p,i,j,a[100005],cnt[1][3],max,increase;
    while(scanf("%ld%ld",&y,&p)!=EOF)
    {
        max=-1;
        increase=-1;
        for(i=0;i<p;i++)
        {
            scanf("%ld",&a[i]);
            if(i!=0&&a[i]<a[i-1])
            {
                increase=0;
            }
        }
        if(increase==0)
        {
            sort(a,a+p);
        }
        for(i=0;i<p;i++)
        {
            cnt[0][2]=0;
            for(j=i;j<p&&a[j]<=a[i]+(y-1);j++)
            {
                cnt[0][2]++;
            }
            if(max<cnt[0][2])
            {
                max=cnt[0][2];
                cnt[0][0]=a[i];
                cnt[0][1]=a[j-1];
            }
        }        
        printf("%ld %ld %ld\n\n",max,cnt[0][0],cnt[0][1]);
    }
    return 0;
}]

thanks...

Posted: **Fri Sep 13, 2013 9:00 am**

Why RE ? Please help

Code: Select all

#include <iostream>
#include<vector>
#include<cstdio>
using namespace std;

int main()
{
    freopen("input.txt", "r", stdin);

    int span;

    //scanf("%d",&span);
     while ( scanf("%d",&span) != EOF ) {
          vector<int> a(100001);
          vector<int> F(100001);

          int Y , year;
          a[0] = F[0] = 0;



        scanf("%d",&Y);

        span -=1;
        for(int i=1; i<=Y; i++)
        {
            scanf("%d",&year);
            a[year] +=1;
        }

        for(int i=1;i<a.size();i++)
        {
            F[i] = (a[i] + F[i-1]);
        }

        int max=0,from , sum;
        for(int i=1;i<F.size();i++)
        {

            if(i+span < F.size())
                sum = F[i+span] - F[i-1];
            else
                sum = F[F.size()-1] - F[i-1];

            if(sum > max)
            {
                max=sum;
                from = i;
            }
        }



        printf("%d %d %d\n", max , from , from+span);

     }
    return 0;
}

Posted: **Fri Sep 13, 2013 9:23 pm**

Don't read from a file.

Posted: **Mon Sep 16, 2013 1:06 am**

@brainfry713 When i'm submitting code i'm removing freopen() [ sorry I did't mension that in code].
Is there any other problem that can end with a RE ?

Posted: **Tue Sep 17, 2013 10:04 pm**

year L <= 1,000,000

Posted: **Sun Oct 05, 2014 11:07 pm**

Hello all,
I'm getting RE and I can't find out the reason of that.
Here's my code:

Code: Select all

Accepted

Appreciate your help

EDIT:
For those who are getting RE notice that P<=100000 in the judge's input (not 5000 as in the problem statement).

Posted: **Tue Oct 07, 2014 10:50 pm**

There is a case in the judge's input where P > 5000, but not one where P > 100000.
http://online-judge.uva.es/board/viewto ... &hilit=957

Posted: **Tue Oct 07, 2014 10:58 pm**

brianfry713 wrote:There is a case in the judge's input where P > 5000, but not one where P > 100000.
http://online-judge.uva.es/board/viewto ... &hilit=957

Thanks a lot. Got Accepted

Posted: **Sat Oct 18, 2014 7:28 pm**

Jan wrote:Try the cases.

Input:
Code: Select all
11
20
20 numbers given in sample :)
12
20
20 numbers given in sample :)
17
20
20 numbers given in sample :)
Output:
Code: Select all
11 12 21
11 12 21
13 6 21
Hope the weird input set helps. .

Could someone tell me how is this correct?
From years 11-21 we had 12 popes.
There was a pope elected in year 8 and he was alive up until year 12 when a new one was elected.
pope 5 was from 8-11, pope 6 in 12, pope 7 and 8 | 13-14, pope 9 15, pope 10 16, pope 11 in 17, in year 21 pope 16.
So, there were popes from 5 to 16 on a period of 11 years. Pope 5 being in year 11, and pope 16 in year 21.
21 - 11 + 1 = 11 years...
Shouldn't the answer be 12 8 21, instead of 11 12 21?

It is obvious that the text of the task is completely unclear. There's no limit between difference of election years of first and last pope.
And somehow all of the answers automatically assume that last-first+1 <= Y.

Posted: **Tue Oct 21, 2014 8:13 pm**

Input:

Code: Select all

Correct output:

Code: Select all

11 12 21
11 12 21
13 6 21

From years 12 to 21 there were 11 popes elected. There was no pope elected in year 11. We are only interested in the years of election, not the number of popes in office during the time frame. The problem statement is misleading.

OJ Board

957 - Popes

957 - Popes

957-Popes

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Re: 957 - Popes

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Re: 957 - Popes

Re: 957 - Popes

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Re: 957 - Popes