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.

What is the WA#5?Какой WA#5?

Можете объяснить, зачем даётся переменная m. Я успешно решил задачу без её использования:

        public static void Main()
        {
            var queryCount = int.Parse(Console.ReadLine().Split(' ')[0]);
            var queries = Enumerable
                .Range(0, queryCount)
                .Select(number => Console.ReadLine())
                .ToArray();
            foreach (var line in Solve(queries))
                Console.WriteLine(line);
        }

Вот код... Я не понимаю как уменьшить расход памяти... но программа написана верно и работает...

#include<iostream>
using namespace std;
int main(){
 int n,i,j;

 float x;
 cin >> n;
 unsigned int a[250000];
 for(i=1;i<=n;i++){
                  cin>>a[i];
                  }
 bool k=true;
 for (i=1;(i<n && k==true);i++){
     k=false;
     for(j=1;j<=n-i;j++){
     if (a[j]>a[j+1]) { swap(a[j],a[j+1]);
     k=true;
     }
     }
     }
     if (n%2==0){
    cout << (a[n/2]+a[n/2+1])/2;
     }
     else {cout << a[n/2];}
}

I solved this problem using a Binary indexed tree modification(read about Counter tree fenwick). It utilizes 2n memory and lets you find MAX of a subarray in logarithmic time.
My solution: 0.062s, 812 kb, yet i haven't tried to do some optimization, I used std::vector which is slower than an array. Nevertheless, I'm pretty ok with such a result

Could someone provide any hint?



Edited by author 10.06.2017 23:14

var
a,b,c,d:integer;
begin
Read(a,b,c,d);
While a<>c do
      begin
      a:=a+b;
      c:=c-d
      end;
a:=c;
WriteLn(a);
end.


Почемууу((
Why(((

Hi. My solution works on my computer but there is not.

What's wrong with my code?

#include <iostream>

using namespace std;

int main()
{
    int n, c = 1;
    int pix[100][100];

    for ( int i = 0; i < n; ++i ) {
        for ( int j = 0; j < n; ++j ) {
            pix[i][j] = 0;
        }
    }

    cin >> n;

    for ( int i = 0; i < n; ++i ) {
        for ( int j = (n - 1); j >= 0; --j ) {
            if ( pix[i][j] > 0 )
                continue;

            int ii =  i, jj = j;

            while ( ii < n && jj < n ) {
                pix[ii][jj] = c;
                ii++;
                jj++;
                c++;
            }
        }
    }

    for ( int i = 0; i < n; ++i ) {
        for ( int j = 0; j < n; ++j ) {
            cout << pix[i][j] << " ";
        }
        cout << endl;
    }

    return 0;
}

Edited by author 10.06.2017 15:16

Edited by author 10.06.2017 15:16

Edited by author 10.06.2017 15:17

import java.util.Scanner;


public class _1206 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = 36;
        long k = s.nextInt();
        while (k > 1) {
            n *= 55;
            k--;
        }
        System.out.println(n);
    }
}

Edited by author 05.05.2014 17:02

var
   a:array [1..200000] of int64;
   n,i: integer;
begin
   while not seekeof  do
   begin
     inc(i);
     read(a[i]);
   end;
   for n:=i downto 1 do
   begin
     writeln(sqrt(a[n]):4:4);
   end;
end.

I copied the solution from the other guy
p.s. Sorry man, i dont know your nickname

Edited by author 09.06.2017 11:43

var i,j,n,n#,n2,a,m:integer;
       l:array[#..400#,#..2] of longint;
begin
read(n);
for i:=# to n do read(l[i,#]);
read(n1);
for i:=# to n# do begin
                              read(a);
                              for j:=# to 400# do if l[j,#]=a then l[j,2]+=#;
                              end;
read(n2);
for i:=# to n2 do begin
                              read(a);
                              for j:=# to 400# do if l[j,#]=a then l[j,2]+=#;
                              end;
for i:=# to 400# do if l[i,2]=2 then m:=m+#;
writeln(m);
end.

Where is # change it with 1

using System;
using System.Collections.Generic;
using System.Linq;

namespace ThreeprimeNumbers
{
    public static class IntExtensions
    {
        public static bool IsPrime(this int number)
        {
            for (var divisor = 2; divisor < number; divisor++)
                if (number % divisor == 0)
                    return false;
            return true;
        }
    }
    internal static class Solver
    {
        private static Func<IEnumerable<int>> ThreeDigitPrimeNumbers =
            () => Enumerable.Range(100, 900).Where(number => number.IsPrime());
        private static Dictionary<int, IEnumerable<int>> PrefixCorrespondance(IEnumerable<int> primeNumbers)
        {
            return primeNumbers
                .GroupBy(number => number / 10)
                .ToDictionary(group => group.Key, group => group.Select(item => item));
        }
        internal static long Solve(int digitCount)
        {
            var table = new long[digitCount - 2, 90];
            var prefixes = PrefixCorrespondance(ThreeDigitPrimeNumbers());
            for (var counter = 10; counter < 100; counter++)
                table[0, counter - 10] = (prefixes.TryGetValue(counter, out var collection)) ? collection.Count() : 0;
            for (var counter = 1; counter <= digitCount - 3; counter++)
                foreach (var pair in prefixes)
                    foreach (var primeNumber in pair.Value)
                    {
                        var suffix = primeNumber % 100 - 10;
                        table[counter, pair.Key - 10] += (suffix < 0) ? 0 : table[counter - 1, suffix];
                        table[counter, pair.Key - 10] %= 1000000009;
                    }
            long sum = 0;
            for (var counter = 0; counter < 90; counter++)
                sum = (sum + table[digitCount - 3, counter]) % 1000000009;
            return sum;

        }
        public static void Main()
        {
            int.TryParse(Console.ReadLine(), out int digitCount);
            Console.Write(Solve(digitCount));
        }
    }
}

I use ferrari method to find roots. Then i got next equation 1/((x-a)*(x-b)*(x-c)*(x-d)), where a,b,c,d -roots of equation. How to calculate such an integral??? If someone can answer here or kilik94@yandex.ru, i would be very grateful!

What is the test data for Test#2?

package main

import (
    "bufio"
    "os"
    "strconv"
    "fmt"
    "strings"
)

func main() {
    reader := bufio.NewReader(os.Stdin)
    firstLine, _ := reader.ReadString('\n')
    count, _ := strconv.Atoi(strings.Trim(firstLine, "\n"))
    result := make([]int, count)

    for i := 0; i < count; i++ {
        position, _ := reader.ReadString('\n')
        result[i] = countFieldsUnderAttack(position)
    }

    for _, val := range result {
        fmt.Println(val)
    }
}

func countFieldsUnderAttack(position string) int {
    result := 0

    var positionInts [2]int
    positionInts[0] = int(position[0]) - int('a') + 1
    positionInts[1] = int(position[1]) - int('1') + 1

    moves := [8][2]int{
        {1, 2},
        {2, 1},
        {-1, -2},
        {-2, -1},
        {-1, 2},
        {2, -1},
        {1, -2},
        {-2, 1},
    }

    for _, move := range moves {
        if move[0] + positionInts[0] > 0 && move[0] + positionInts[0] <= 8 {
            if move[1] + positionInts[1] > 0 && move[1] + positionInts[1] <= 8 {
                result++
            }
        }
    }

    return result
}

Edited by author 08.06.2017 23:12