>

>

> 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

We use 'eof' in pascal.

My code:
procedure do;
var     n:int64;
begin
        if eof then exit;
        read(n);
        do;
        writeln(sqrt(n):0:4);
end;

begin
        do;
end.



oh I see!

http://acm.timus.ru/forum/thread.aspx?space=1&num=1001&id=23490&upd=633934535278966250

Edited by author 18.07.2010 19:15

How can we use eof without opening file?? WTH with it? I can't understand anything! FPC 2.0 doesn't compiling it. And Embarcadero RAD Studio XE too! So, What The Hell with it?

the graph may be not connected.

I got WA#2 too,
but I thought that the graph is supposed to be not connected.

Can help me identify my errors , here's my solution :
- I reverse all the edges
- DFS from the vertex without any other vertex pointing to
- Count the depth, if depth = number of vertex -> output in reverse order

3Q,i made the same mistake,and i now get AC

Timus use the compiler visual c++ compiler not gcc.
hint: log2(n) = log(n)/log(2)
(but this sum at the complexity)

Edited by author 21.01.2011 22:32

Edited by author 21.01.2011 22:32

Timus use the compiler visual c++ compiler not gcc.
hint: log2(n) = log(n)/log(2),
(but this sum at the complexity)