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

#include <iostream>
using namespace std;
int main()
{
    int a,b;
    cin >>a;
    if (a>=1){
        for (int i=1;i<=a;i++){
            b=b+i;
        }
    }
    else if (a<1){
        for (int i=1;i>=a;i--){
            b=b+i;
        }
    }
    cout<<b;
}


i tested with test case #1 it work, but after submit the judge says wrong answer test 1, why?

Make sure that:
Ki (the second input line) is given in reverse order:
eg. Kn, K(n-1), ... K2, K1.
But, Ci (the third input line) is not:
eg. C1, C2, ..., Cn.
In the sample test, this will not make you sence...

This is not declared in statement.
Admin: will you add this?

Edited by author 02.10.2010 21:20

Edited by author 02.10.2010 21:20

Don't use 'string' type.

8m3edk-w4h74s:1:9: fatal error: 'iostream.h' file not found
#include<iostream.h>
        ^
1 error generated.

Is first test from problem description? I can't find any issue

Find maximal matching using Kuhn algo.  Then greedily add missing connections.
That is slow way.
My C++ solution without any STL use runs in 230-240 ms.

#include <iostream>
#include <iomanip>
#include <cmath>
using std::cin;
using std::cout;
using std::endl;
using std::setprecision;
using std::fixed;
int main()
{
    int n;
    double r,s=0,x,y,x0,y0,xn,yn;
    cin >> n >> r;
    if (n==1) {
        cin >> x >> y;
    }
    if (n>1) {

            cin >> x >> y;
    for (int i=1;i<=n-1;i++) {
    cin >> x0 >> y0;
    s=s+sqrt((x0-x)*(x0-x)+(y0-y)*(y0-y));
    x=x0;y=y0;
    }
    s=s+sqrt((xn-x)*(xn-x)+(yn-y)*(yn-y));
            }
s=s+2*3.1415*r;
cout << fixed << setprecision(2) << s << endl;

    return 0;
}

DEV-C++ 5.11

Edited by author 04.06.2017 12:53

any hints plzzzz

C++ Clang 14.

Can someone tell me what the test#13 example please?

Я долго бился с этой проблемой несмотря на то,что сложность <200. Я пытался решить с помощью ДП,но там очень много случаев и сложная реализация. Задачу можно решить логически: во-первых,первые символы "<" и последние символы ">" ни на что не влияют,поэтому удалим их; во-вторых,если слева есть хоть один ">",то он в ЛЮБОМ случае будет меняться с КАЖДЫМ "<" РОВНО один раз(если "<" существует). Т.е. нас не интересует конструкция строки. Нас интересуют числа Z[1],E[1],Z[2],E[2],...Z[p],E[p],где Z[i] -количество ">" в последовательности 00000000...00000,а E[i]- количество "<" в последовательности 11111....11111111. Т.е. нашу строку делаем в виде 00...0011..1100..0011..11........00..0011..11,где P чередований (я ">" заменил на "0",а "<" на "1")
. Тогда ответ есть Z[1]*(E[1]+...E[p])+Z[2]*(E[2]+..+E[p])+...+Z[p]*E[p]. Сумму можно посчитать через частичные с помощью ДП. Асимптотика O(N).

The problem is quite evil.  The problem is actually very easy.  It is hard to understand that the problem is that easy.
Possible solution.
First ,  eliminate all such segments I=[l;r],  that l>M or r<0, because of such I's can't appear in the answer.
Second,  build your dp[].
Your dp[x]  contains rightmost right end of all such segments [l;r],  that l<=x.
Finally,  just start from segment whose right end is dp[0],  jump to a  segment whose right end is dp[dp[0]] and so on.

How to build dp[x]?
I have used an additional array right_end[y].  Where right_end[y] is the right end of a segment with maximal length from all segments whose left end is y.
dp[x] =max(dp[x-1], right_end[x]).

What does it mean? Runtime error (access violation)
This is my code on c++
#include <iostream>
using namespace std;
int n,ans;
string s;
int main()
{
    cin>>n;
    ans=n+2;
    for(int i=0; i<n; i++)
    {
        cin>>s;
        for(int j=0; j<s.size()-3; j++)
        {
            if(s[j]=='+' && s[j+1]=='o' && s[j+2]=='n' && s[j+3]=='e')
            {
                ans++;
                break;
            }
        }
    }
    if(ans==13) ans++;
    cout<<100*ans;
    return 0;
}

Edited by author 31.05.2017 21:40

I used the same code