Why if I am trying to precalculate factorials using unsigned long long int there is WA#5?  There are 64 last binary  digits  digits and in this task we are to output at most 32 binary digits.

There are no tests where N=0 xD)

#include<cstdio>
#include<cstring>
#include<iostream>
#define pii pair<int,int>
#define pdd pair<double,double>
#define mp make_pair
#define F first
#define S second
#define N 210
using namespace std;
int n,m;
int fat[N];
pii e[N];
pdd p[N];
int father(int x) {if(fat[x]!=x) fat[x]=father(fat[x]); return fat[x];}
double calc(pii a,pii b,pii c)  {return (a.F-c.F)*(b.S-c.S)-(b.F-c.F)*(a.S-c.S);}
bool judge(pii a, pii b, pii c, pii d)
{
    if (max(a.F,b.F)<min(c.F,d.F)||max(a.S,b.S)<min(c.S,d.S)||max(c.F,d.F)<min(a.F,b.F)||max(c.S,d.S)<min(a.S,b.S))  return false;
    if (calc(c,b,a)*calc(b,d,a)<0||calc(a,d,c)*calc(d,b,c)<0)  return false;
    return true;
}
int main()
{
    int i,j,x,y;
    scanf("%d %d",&n,&m);
    for(i=1;i<=n;i++) fat[i]=i;
    for(i=1;i<=n;i++) {scanf("%lf %lf",&x,&y); p[i]=mp(x,y);}
    for(i=1;i<=m;i++) {scanf("%d %d",&x,&y); e[i]=mp(x,y); fat[father(x)]=father(y);}
    for(i=m+1;i<=m+n;i++) e[i]=mp(i-m,i-m);
    m+=n;
    for(i=1;i<=m;i++)
        for(j=i+1;j<=m;j++)
            if(judge(p[e[i].F],p[e[i].S],p[e[j].F],p[e[j].S]))
                fat[father(e[i].F)]=father(e[j].F);
    int stan=father(1);
    for(i=1;i<=n;i++) if(father(i)!=stan)
    puts("YES");
    return 0;
}

Just output n random numbers.
After that,  output m random numbers.
All random numbers are in range [1;1e9].
AC first attempt.

At first I wrote a tricky solution using trigonometry and binary search.  And I was unable to pass through test case No.  12.
Then I wrote simple two-step solution and got AC.

Can somebody write input?

I wrote program on C++ and got WA1. I rewrote program on Pascal and got AC.
I think my error in input string. How do i read lines? My attempts:
1)
     int c;
     n=0;
     while ((c = getchar()) != EOF)
     {
          if(c=='\n') break;
          s[++n]=c;
     }
     m=0;
     while ((c = getchar()) != EOF) t[++m]=(char)c;
2)
void Input()
{
     int c;
     scanf("%c",&c);
     n=0;
     while (c!='\n')
     {
          s[++n]=c;
          scanf("%c",&c);
     }
     m=0;
     while ((c = getchar()) != EOF) t[++m]=(char)c;
}

Probably <string> will help, but i prefer work with index>=1.

Edited by author 22.06.2017 23:44

Edited by author 22.06.2017 23:44

Used DP + BIT, AC in 0.078
O(n*k*logn) solution.
Don't forget, % 10^9 every step.

My approach wrong approach is such.
Read cur number X.
Find CNT count of numbers Y (Y>X) already read.
Find number of ways to pick K-1 items from set of CNT distinctive  items.
Add that number to ANSWER. Find modulo.

Edited by author 11.05.2017 13:28

After we add new pile Z of stones if there is new pair of piles X and Y such that |X-Y| is minimal that either X=Z or Y=Z. So,  just add nee pile and recalculate new pair of piles such that the difference between stones count is minimal in O(n)  where n is tje number of piles present now.

import java.util.Scanner;

/**
 * Created by macbookpro on 2/19/15.
 */
public class JavaThieves877 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int c1 = Integer.parseInt(scanner.nextLine());
        int c2 = Integer.parseInt(scanner.nextLine());

        if(c1 % 2==0 || c2 % 2!=0 )
        {
            System.out.print("Yes");
        }
        else if(c1 % 2 != 0 || c2% 2== 0)
        {
            System.out.print("No");
        }
    }
}

Some tests?

At first I thought I was getting WA's because of the wrong minus sign.
Actual reason was printing integers.
I was trying to print integers using just PUTCHAR.
Of course it worked fine for 0-9. And was failing for other numbers.

I tried to solve it with C++.  It was long.  It was PAINFUL!  Finally,  I didn't managed to solve it in C++.  And then I just opened my Python IDE and solves it within 10 minutes.  Just AC first attempt.  The task is evil.

The product of natural numbers with a given condition in this problem will be maximal if and only if the middle arithmetic of these numbers is the most closely to the number e=2.71828... It's a mathematical sentence which are proved!

P.S. Sorry for my bad English

used RMQ using segment tree to find lca in log(n)
total time complexity O(n+q*(lgn)) = O(q*lgn)

Some graph problems having very low rating (caravans,  Ivan's car)  are quite hard to solve while some other problems (like Gear-wheels,  Labyrinth)  have rating two times higher and are incredibly easy.

For each x (x=0,1,2,..,n-1) (x is horizontal axis)  find y0=x*tg(a) and y1=(x+1)*tg(a),  where tg(a) is (m-1)/(n-1).  Add to answer floor (y1) - floor(y0) +1. If y1 is a whole number then subtract 1 from answer.

program pr;
     Var
     n,k,i,s,q:integer;
     a:array[1..100000] of integer;
     b:array[1..100000] of integer;
Begin
     q:=0;
     read(n,k);
     for i:=1 to n do
     readln(a[i]);
     for i:=1 to n do
     b[i]:=k-a[i];
     for i:=1 to n do begin
     if b[i]<0 then s:=abs(b[i]);
     if b[i]>=0 then q:=q+b[i] ;
     end;

write(s,' ',q)
End.