some tests, that could help you

1
0 0
anwear:
*


3
0 0
0 5
5 0
answear:
*.....
|.....
|.....
|.....
|.....
*----*


3
0 1
0 5
0 9
answear:
*
|
|
|
*
|
|
|
*
+

Hi, your checker doesn't check that exactly M lines are printed. My submission #5318694 has received AC.

Can someone give me an input example with the right output so I could check it, 'cause all the test I run are right

i use binsearch and have wa 15..

i found this test
1 1 5 1 2
what's answer? is this test correct?

point is equal.
e.g.
3 3
1 1
1 1
111
111
111

answer:1 0

Can someone give me data of test #11? I have test all data in discussion section and got right answer but my program just stuck in test #11....

I wrote a simple solution (http://acm.timus.ru/getsubmit.aspx/5309080.go), and it got WA3. The second solution (http://acm.timus.ru/getsubmit.aspx/5309073.go) got AC. I can't realise why v1 reads test #3 incorrectly. Is it a fmt-library error?

Some new tests were added. 234 authors lost AC. Maybe we'll add more tests soon.

Каким должен быть ответ ,если значения n или k не валидны?
Например, если не указать k. Или ввести значения, не попадающие в диапазон?

I don't know why I get WA on test 4!
Can there be a negative number?
This is my code:

#include<iostream>
#include<cmath>
using namespace std;
long long dyn[110][510] , a[110][510] , big, sum[110][510] , p[110][510][2] , o , m, n;
int main(){
    cin>>m>>n;
    for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
           cin>>a[i][j];
    for(int i=0;i<m;i++)
    {
        sum[i][0]=a[i][0];
        dyn[i][0]=100000000000000000;
        for(int j=1;j<n;j++)
        {
            sum[i][j]=sum[i][j-1]+a[i][j];
            dyn[i][j]=100000000000000000;
        }
    }
if((m==1)&&(n==1))
{
cout<<1<<endl;
return 0;
}
    dyn[m-1][0]=a[m-1][0];
    for(int i=1;i<n;i++)
    {
        dyn[m-1][i]=a[m-1][i];
    }
    for(int i=m-2;i>=0;i--)
    {
        for(int j=n-1;j>=0;j--)
        {
            for(int k=0;k<n;k++)
            {
                if(j>k)
                {
                    if(dyn[i+1][k]+sum[i][j]-sum[i][k-1]<dyn[i][j])
                    {
                        dyn[i][j]=dyn[i+1][k]+sum[i][j]-sum[i][k-1];
                        p[i][j][0]=j;
                        p[i][j][1]=k;
                    }
                }
                else
                {
                    if(dyn[i+1][k]+sum[i][k]-sum[i][j-1]<dyn[i][j])
                    {
                        dyn[i][j]=dyn[i+1][k]+sum[i][k]-sum[i][j-1];
                        p[i][j][0]=j;
                        p[i][j][1]=k;
                    }
                }
            }
        }
    }
    big=dyn[0][0];
    o=0;
    for(int i=1;i<n;i++)
    {
        if(dyn[0][i]<big)
        {
            big=dyn[0][i];
            o=i;
        }
    }
    cout<<o+1;
    for(int i=1;i<m-1;i++)
    {
    if(p[i][o][0]<=p[i][o][1])
    {
        for(int j=p[i][o][0];j<=p[i][o][1];j++)
            cout<<" "<<j+1;
            }
            else
            {
            for(int j=p[i][o][0];j>=p[i][o][1];j--)
                cout<<" "<<j+1;
                }
            o=p[i][o][1];
    }
    cout<<" "<<o+1<<endl;
return 0;
}

Хотелось бы пример теста
#pragma comment(linker, "/STACK:16777216")

#include <iostream>
using namespace std;

char  str[100001];

bool isPalind(int start, int finish)
{
    int fin = finish-1;//(int)finish-48 - 1;
    bool answer = true;
    for (int k=start-1; k<fin; k++)
    {
        if (str[k] != str[fin])
        {
            return false;
        }
        fin--;
    }
    return true;
}



int main()
{
    char c1, c2;
    int pos=0, m, fin;
    cin >> str;
    scanf("%d\n", &m);
    for (int i=0; i<m; i++)
    {
        scanf("%*[^? ]%c%d ", &c1, &pos);
        if (c1 == ' ')
        {
            scanf("%c", &c2);
            str[pos-1] = c2;
        }
        if (c1 == '?')
        {
            scanf("%d", &fin);
            if(isPalind(pos, fin))
            {
                printf("Yes\n");
            }
            else
            {
                printf("No\n");
            }
        }
    }
    return 0;
}

Edited by author 01.11.2013 00:11

What it is about?

How I can do it on 0.001?

Edited by author 31.10.2008 02:29

ATT

I don't see any dynamic programming in this problem...

Edited by author 02.11.2013 13:47

I was trying many times to get past first test and fixed a few mistakes in the algorithm to no avail.
And the reason to fail was that the output was in brackets.

У меня все работает с разными данными. Мне кажется я не правильно поток ввода установил. Здесь нужно использовать cin?

How it is posible to get AC in 0.001 s? What time do such ACs take to get L for K= 99999989?
Could anyone give a hint to make it faster than O(sqrt(K))?