Oh, now I see.. Admins, please correct russian statement.
Look at this:

English: If there are many such schemes, then you should choose an optimal scheme among them, i.e., a scheme that also reduces sequences of up to K spaces (K ≥ L) for a maximal possible K. If there are several optimal schemes, you may give any one of them.

Russian: Если таких планов несколько, то вы должны выбрать среди них оптимальный — такой, который позволяет заменить последовательности пробелов длины K (K ≥ L) для максимально возможного K.

These conditions are not equal! An example you can see in my previous post.

Good Luck!

I have the same trouble. But after some observings i found that i misunderstood statement. We need to find distance between resulting CENTER of the square and the given point.

I advice you to read books of Martin Gardner. There is beautiful solution of this problem. Or you can write DP and look at the points where stations are built.

Eh... You can use "scanf()==EOF" or "(cin>>...)!=0", which may not work expectedly in your own machine.

Edited by author 27.01.2011 12:26

Isn't there any "Stack Overflow" error occured? And how can you terminate the input? I have tried "scanf(...)==EOF" and (cin>>...)!=0, but both failed.

I got it...

The size of the edges is too small...

It's RE...

>

>

> Who knows, who knows...

6700

Tank you very much

Thank :-)!

Thank you!

I love you.

A got AC using real numbers with EPS = 1e-8, so precision is not major matter here...

Thanks for idea. Simple and beautiful. My previous AC solution was more difficult.

Yes, random work. My solution ~ 0.5s. Choose three random points and count points inside circle. Maybe weak tests...

Edited by author 27.01.2011 00:28

If you have a group with N spaces and apply to it the operation, that changes K spaces to one, then the number of spaces will be N/K + N%K. You should output such numbers K[1], K[2], ... , K[m], that for every number N in range [1..L] a sequence of operations N := (N/K[i] + N%K[i]) for each i = 1..m will make the value N equals 1. And then you should maximize the range for N. I guess it is a school problem :)

Edited by author 25.01.2011 20:42

Edited by author 25.01.2011 20:43

My code:

#include <stdio.h>

void printdw(int* f1, int *f3, int d, int dd, int lim)
{
    int f2=0;
    if (d==1 || *f1==1)
    {
        if (dd==d) {printf(" [%2d]", d); f2=1;}
        else {printf("  %2d", d);}
        *f1=1;
    }
    else
    {
        if (dd==d) {printf("      [%2d]", d); f2=1; *f1=1;}
        else {printf("       %2d", d);}
    }
    for(d+=7; d<=lim; d+=7)
    {
        if (f2==1) {printf("  %2d", d); f2=0;}
        else
        {
            if(d==dd) {printf("  [%2d]", d); f2=1;}
            else printf("   %2d", d);
        }
    }
    if (d==lim+7)
        *f3=1;
    if(d<lim+7 && *f3) { if (!f2) printf("     "); else printf("    ");}
    printf("\n");
}

int main()
{
    int dd, mm, yy, dw, lim, f1, f2, f3, d;
    scanf("%d %d %d", &dd, &mm, &yy);
    dw=(6+yy-1600+((yy-1)/4-399)-((yy-1)/100-15)+((yy-1)/400-3))%7;
    if (mm==5) dw++;
    if (mm==8) dw+=2;
    if (mm==2 || mm==3 || mm==11) dw+=3;
    if (mm==6) dw+=4;
    if (mm==9 || mm==12) dw+=5;
    if (mm==4 || mm==7) dw+=6;
    if (mm>2) { if (!(yy%4)) dw++; if (!(yy%100)) dw--; if (!(yy%400)) dw++;}
    dw%=7;
    if (mm==1 || mm==3 || mm==5 || mm==7 || mm==8 || mm==10 || mm==12) lim=31;
    else
    {
        if (mm!=2) lim=30;
        else {if (yy%4 ||(!(yy%100) && yy%400)) lim=28; else lim=29;}
    }
    f1=0; f3=0;
    printf("mon"); d=(8-dw)%7+1;
    printdw(&f1, &f3, d, dd, lim);
    printf("tue"); d=(d+7)%7+1;
    printdw(&f1, &f3, d, dd, lim);
    printf("wed"); d=(d+7)%7+1;
    printdw(&f1, &f3, d, dd, lim);
    printf("thu"); d=(d+7)%7+1;
    printdw(&f1, &f3, d, dd, lim);
    printf("fri"); d=(d+7)%7+1;
    printdw(&f1, &f3, d, dd, lim);
    printf("sat"); d=(d+7)%7+1;
    printdw(&f1, &f3, d, dd, lim);
    printf("sun"); d=(d+7)%7+1;
    printdw(&f1, &f3, d, dd, lim);
}

Please! help to find mistake!

Edited by author 24.01.2011 17:54