c++:


#include<iostream>
using namespace std;

long long sqr(int x, int y){
    long long a=1;
    for(int i=1;i<=y;i++)
            a*=x;
    return a;
}

int main(){
    int n,y,m,a=-1;
    cin>>n>>m>>y;
    if(y>=m){
           cout<<-1<<endl;
           return 0;
    }
    for(int x=0;x<m;x++)
            if((sqr(x,n))%m==y){
                                cout<<x<<" ";
                                a+=2;
            }
    cout<<endl;
    if(a==-1 && y<m)
             cout<<-1<<endl;
    return 0;
    }

please help me!

Edited by author 20.01.2011 20:31

Edited by author 20.01.2011 20:33

Reading by Console.Readline() or Console.ReadToEnd gets Memory Limit, reading by Console.Read() gets Time Limit.
Is there any other way to read without getting MLE and TLE?

Hint is wrong. I had WA#15, but after changing in Java code "int" to "long" task was accepted.

# include <stdio.h>
# include <algorithm>

using namespace std;

struct daw
{
    long long rg,lf,nd,vl,mx;
};

long long a,b;

long long w[409];
long long c[409][5];
long long v[409][409];

daw d[409];

void btr(long long x)
{
    if(d[x*2+1].vl!=0)
    btr(x*2+1);

    if(d[x*2+2].vl!=0)
    btr(x*2+2);

    if(x!=0)
    d[(x-1)/2].nd=d[(x-1)/2].nd+d[x].nd+1;
}

long long dp(long long x, long long y)
{
//    prlong longf("%d %d",x,y);getchar();

    if(y==0)
    return 0;

    if(v[x][y]>0)
    return v[x][y];

    for(long long i=0;i<=y;i++)
    if(d[x].nd-d[x*2+2].nd>=i && d[x*2+1].vl>0 && d[x*2+2].vl>0)
    {
        long long q=0;

        if(i!=0)
        q+=d[x].lf;

        if(i!=y)
        q+=d[x].rg;

        if(i==0)
        v[x][y]=max(v[x][y],dp(x*2+2,y-1)+q);

        else if(i==y)
        v[x][y]=max(v[x][y],dp(x*2+1,y-1)+q);

        else
        v[x][y]=max(v[x][y],dp(x*2+1,i-1)+dp(x*2+2,y-i-1)+q);
    }

    return v[x][y];
}

int main()
{
    scanf("%lld %lld",&a,&b);

    for(long long i=0;i<a-1;i++)
    scanf("%lld %lld %lld",&c[i][0],&c[i][1],&c[i][2]);

    d[0].vl=1;
    w[1]=1;

    for(long long i=0;i<a-1;i++)
    for(long long j=0;j<a-1;j++)
    {
        if(d[i].vl==c[j][0] && w[c[j][1]]==0)
        {
            if(d[i*2+1].vl==0)
            {d[i].lf=c[j][2];  d[i*2+1].vl=c[j][1];}

            else
            {d[i].rg=c[j][2];  d[i*2+2].vl=c[j][1];}

            w[c[j][0]]=1;
        }

        if(d[i].vl==c[j][1] && w[c[j][0]]==0)
        {
            if(d[i*2+1].vl==0)
            {d[i].lf=c[j][2];  d[i*2+1].vl=c[j][0];}

            else
            {d[i].rg=c[j][2];  d[i*2+2].vl=c[j][0];}

            w[c[j][0]]=1;
        }
    }

    btr(0);

    if(b==1)
    {printf("%lld",max(d[0].rg,d[0].lf));return 0;}

    printf("%lld",dp(0,b));

    getchar();
    getchar();
}


Thanks

[edit]  I used formula for day of week instead of pascal function. Got ac.
I am using dayofweek to get the weekday for the first of the month in the question.
I am getting WA7.
I think my formatting is OK but I am getting wrong answers for some inputs.
The two supplied examples are OK, and
20 10 2050
mon        3   10   17   24   31
tue        4   11   18   25
wed        5   12   19   26
thu        6   13  [20]  27
fri        7   14   21   28
sat   1    8   15   22   29
sun   2    9   16   23   30

and
25 11 1885
mon        2    9   16   23   30
tue        3   10   17   24
wed        4   11   18  [25]
thu        5   12   19   26
fri        6   13   20   27
sat        7   14   21   28
sun   1    8   15   22   29

but not
4 7 1890
mon      [ 4]  11   18   25
tue        5   12   19   26
wed        6   13   20   27
thu        7   14   21   28
fri   1    8   15   22   29
sat   2    9   16   23   30
sun   3   10   17   24   31

or
1 1 1600
mon [ 1]   8   15   22   29
tue   2    9   16   23   30
wed   3   10   17   24   31
thu   4   11   18   25
fri   5   12   19   26
sat   6   13   20   27
sun   7   14   21   28

from what I have seen in the forum.

Any ideas please?
Thanks




Edited by author 05.12.2012 22:33

var
a,b:integer;
begin
readln(a,b);
writeln(a+b);
end.

"highly protable"? May be "highly profitable"?
And "in those aairs"? May be "in those affairs"?

Edited by author 05.12.2012 10:26

Firstly i calculate the longest rope lenths at each tree.
Then just try to tether goats to every tree, calculating the square.
If sum of 2 maxrope are larger then lenght between trees:
double maxrad = std::max(maxr[i], maxr[j]);
double minrad = l - maxrad;
if (minrad < 1.0)
{
    maxrad = maxrad - (1 - minrad);
    minrad = 1.0;
}
if (maxrad >= 1.0)
maxsize = std::max(maxsize, square(maxrad)+sqree(minrad));

Help me!, I got WA 5!.

Give me some tests

I solved using too mush time
Can anyone explain me how some people did it in 0.031??
Please)

C++, 0.015 sec, 196 Kb

#include <iostream>
using namespace std;

int main()
{
    int n,k;
    cin>>n>>k;
    int m=n/k + ((n%k==0) ? 0:1);

    int * A=new int [m*k];
    for (int i=0;i<n;i++)
    {
        cin>>A[i];
    }
    for (int i=n;i<m*k;i++)
        A[i]=-1;

    for (int i=0;i<m;i++)
    {
        for (int j=0;j<k;j++)
        {
            int h=j*m+i;
            if (A[h]<0)
                cout<<"    ";
            if (A[h]>-1 && A[h]<10)
                cout<<"   "<<A[h];
            if (A[h]>9 && A[h]<100)
                cout<<"  "<<A[h];
            if (A[h]>99 && A[h]<1000)
                cout<<" "<<A[h];
        }
        cout<<endl;
    }

    return 0;
}

Waiting of comments

What's incorrect?????

int main()
{
    int n, k;
    scanf("%d %d", &n, &k);

    vector<int> mas(100);

    mas.assign(100, -1);
    for (int i=0; i<n; i++)
        scanf("%d", &mas[i]);
    int s=n/k;
    if(n%k!=0) s++;

    for (int i=0; i<s; i++)
    {
        for(int j=0; j<k; j++)
            if (mas[s*j+i]!=-1)
            printf("%4d", mas[s*j+i]);
        printf("\n");
    }

  return 0;
 }

Edited by author 26.04.2011 20:30

100 1
1 2 3 4 .. 99 100

ans:
1 2 3 4 .. 99 100


Edited by author 09.07.2011 17:54

i think there must be some stupid mistake.My all tests pass it. please provide me good tests or please,kindly let me know the problem in the code:

[code deleted]

AC at last(.187 sec 121 kb)

very very stupid mistake.
if(x==n-x) just print x not x and n-x :P

Edited by author 12.07.2010 10:41

Edited by author 13.07.2010 18:43

Can someone give me some tests?
I have WA#10, I don't know what else I can fix to solve this problem.
I used hashes, but I think that my simple algorithm can have collisions, because I make hash mod 10000000;
//sorry for my english.
also I changed algorithm, now I don't use mod. But, I still have WA#10.
I used stack + hash, but, all the same

Edited by author 03.12.2012 19:10

Edited by author 03.12.2012 19:34

Where's my mistake?


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

int main()
{
    double n, s;
    cin >> n;
    s = (1 + n) * (abs(n / 2.) + (n < 1));
    cout << s;
    return 0;
}

Is it necessary to make a complete graph.
Is it possible to find the solution in a more easy way?
I always get stuck on #4.