There is no test where A, B and C are on the same line and point A is between B and C  (or C is between A and B). These are pretty tricky cases, maybe add them?

Edited by author 21.01.2017 18:51

hello i have problem with my code and i dont know where is it T_T
here is my code :

var a:real;
i,k:integer;
z:array[1..100]of real;
begin
k:=1;
while not eof do begin
read(a);
z[k]:=sqrt(a);
k:=k+1;
end;
for i:=k-1 downto 1 do begin
    writeln(z[i]:0:4);
    end;
end.

Noise() - is constant, or linear, trigonometric, or random function???

#include <iostream>
#include <stdio.h>
#include <memory.h>

#define NMAX 37
using namespace std;

long cnt[NMAX];
int n;

void genSum(int k, int sum);

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "rt", stdin);
    ///freopen("output.txt", "wt", stdout);
    #endif

    scanf("%d", &n);

    genSum (0 , 0);

    int answer = 0;
    for (int it = 9 * n / 2; it >= 0; --it)
    {
        answer += cnt[it] * cnt[it];
    }

    printf("%d\n", answer);

    return 0;
}

void genSum(int k, int sum)
{
    if (k == n / 2)
    {
        ++cnt[ sum ];
    }
    else
    for (int digit = 0; digit <= 9; ++digit)
    {
        genSum(k + 1, sum + digit);
    }
}

My code is giving WA on test1.
 But it is giving correct ans for all these:-

50000 300 300
422791438

49999 299 299
505277419

25000 269 222
716603086

Can anyone provide more test cases?

2
1 10

My output is: 2 1 2 2 2 2 2 2 2 2 2

Is this corret or Did I understand task wrong..?

Други, кто-нибудь знает что это за несчастный тест? Проверил программу ВРУЧНУЮ в экселе для всех 50 значений "n", и всё сошлось. What's the problem?

[code deleted]

Edited by moderator 19.11.2019 22:58

Edited by author 17.01.2017 16:15

I'm not getting the bug in the code :(. It would be helpful if someone pointed the mistake.

key = {'i' : 1, 'j' : 1,
       'a' : 2, 'b' : 2, 'c' : 2,
       'd' : 3, 'e' : 3, 'f' : 3,
       'g' : 4, 'h' : 4,
       'k' : 5, 'l' : 5,
       'm' : 6, 'n' : 6,
       'p' : 7, 'r' : 7, 's' : 7,
       't' : 8, 'u' : 8, 'v' : 8,
       'w' : 9, 'x' : 9, 'y' : 9,
       'o' : 0, 'q' : 0, 'z' : 0
      }

while True:
    n = raw_input()
    if(int(n) == -1):
        break
    l = int(raw_input())
    d = {}
    ans = []

    for a0 in xrange(l):
        temp = raw_input()
        d["".join([str(key[i]) for i in temp])] = temp

    i = len(n)
    while(i>0):
        if n[0:i] in d.keys():
            ans.append(d[n[0:i]])
            n = n[i:]
            i = len(n)+1
        i-=1

    print "No solution." if len(n) != 0 else " ".join(ans)

Народ, кто может указать на часть моего кода которая тормозит алгоритм? Чистил как мог, не смог выйти за 0.125 секунд

import java.util.Scanner;
public class Timus {
        public static void main (String[] args)
        {
            Scanner in = new Scanner(System.in);

            byte m = in.nextByte(), n = in.nextByte();

            System.out.println( (m*n-1)%2==0 ? "[second]=:]" : "[:=[first]");

            in.close();

        }
}

If the matrix size is 3, with the given contraints, it won't be "2 special cells", because 8 of them will be walls, and will be only one free cell. Could be that those special cells were located at the same place? Please drop some light on that. Thanks

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
    class Program
    {
        static void Main(string[] args)
        {
            int k;
            int n;
            do
            {
            do{
             n = int.Parse(Console.ReadLine());
            } while (!(n >= 2));
            do
            {
               k = int.Parse(Console.ReadLine());
            } while (!(k >= 2 && k <= 10));
            } while (!(n + k <= 18));
            int start = (int)Math.Pow(10, n - 1);
            int end = 0;
            int check = start;
            int BigCount = 0;
                for (int i = n - 1; i >= 0; i--)
                    end += (k - 1) * (int)Math.Pow(10, i);
                int h = end - start + 1;
                int o = 0;
                for (int l = 0; l < n; l++)
                {
                    o += h % 10 * (int)Math.Pow(k, l);
                    h = h / 10;
                }
                for (int i = 0; i < o; i++)
                {
                    int count = 0;
                    double p = check;
                    for (int j = 0; j < n / 2; j++)
                    {
                        if ((p % 10 + p / 10 % 10) > 0) count++;
                        p = p / 100;
                    }
                    if (count != 0) BigCount++;
                    check++;
                    count = 0;
                    p = check;
                    int r = 0;
                    int m = 0;
                    for (int l = 0; l < n; l++)
                    {
                        if (p % 10 == k)
                        {
                            count++;
                            m += (10 - k) * (int)Math.Pow(10, l);
                            r = 10 - k;
                        }
                        p += r;
                        p = (int)p / 10;
                    }
                    check += m;
                }

                Console.Write(BigCount);
            }
        }
Норм решение?

Edited by author 16.01.2017 04:29

Edited by author 16.01.2017 04:30

This is my code:

#include <iostream>
#include <cmath>

using namespace std;
int main()
{
    int a[101], n, x, dis1 = 0, dis2 = 0, lb, rb;
    bool flag = false, flag2 = false;

    cin >> n >> x;

    if ((x == 0))
      flag2 = true;
    else
    {
      if ( n == 0 )
        flag = true;
    }

    for ( int i = 0; i < n; i++ )
    {
      cin >> a[i];
      if ( (x > 0) && (a[i] < x) && (a[i] > 0) )
        flag = true;

      if ( (x < 0) && (a[i] > x)  && (a[i] < 0) )
        flag = true;
    }

    if (flag)
      cout << "Impossible" << endl;
    else
    {
      lb = a[0];
      rb = a[0];
      for ( int i = 1; i < n; i++ )
      {
        if ( (a[i] > 0) && (a[i] > rb) && (a[i] > x) )
          rb = a[i];

        if ( (a[i] < 0) && (abs(a[i]) < abs(lb)) && (a[i] < x) )
          lb = a[i];
      }

      if ( x > 0 )
      {
        dis1 = x;
        dis2 = (int) abs(2*lb) + x;
      }
      else
      {
        dis1 = 2*rb + (int) abs(x);
        dis2 = (int) abs(x);
      }

      if (flag2)
        cout << "0 0" << endl;
      else
        cout << dis1 << " " << dis2 << endl;
    }

    system("PAUSE");
    return 0;
}

---------------------------
I can't understand WHY WA#4
Please, help!!!

My Code perfect AC  with Visual Studio 2013, G++ 4.9, G++ 4.9 C++11.
But Clang 3.5 gives output limit exceeded, or access violation....
Submit codes
7217044
7217043

Some bugs in the validator were fixed. The problem was rejudged. 299 WA submits turned to AC and 84 AC submits turned to WA. If you will find new bugs, please, write to timus_support (at) acm.timus.ru.

Edited by author 09.09.2010 10:57

using namespace std;
int main()
{
    int n, d[13];
    scanf("%d", &n);
    d[0] = 5;
    d[1] = 21;
    d[2] = 12;
    d[3] = 2;
    d[4] = 1;
    d[5] = 4;
    d[6] = 6;
    d[7] = 1;
    d[8] = 4;
    d[9] = 4;
    d[10] = 1;
    d[11] = 0;
    d[12] = 1;
    d[13] = 1;
    printf("%d", d[n]);
    return 0;
}

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <queue>

#define f first
#define s second

#define ll long long
#define ld long double
#define ull unsigned long long

#define pb push_back
#define ppb pop_back
#define mp make_pair

#define sz(x) (int) x.size()
#define all(x) x.begin(), x.end()
#define bit(x) __builtin_popcountll(x)
#define sqr(x) ((x) * 1LL * (x))

#define nl '\n'
#define ioi exit(0);

#define NeedForSpeed ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

#define _7day "IOI2017"

using namespace std;

typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ld, ld> pdd;
typedef pair <ll, ull> hack;

const int N = 1e5 + 7, MxN = 1e6 + 7, mod = 1e9 + 7, inf = 1e9 + 7;
const long long linf = (ll)1e18 + 7;
const long double eps = 1e-15, pi = 3.141592;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1}, dy[] = {0, 1, 0, -1, 1, -1, 1, -1};


    int n, x;
    priority_queue <unsigned int> q;
int main(){
    #ifndef _7day
        freopen (_7day".in", "r", stdin);
        freopen (_7day".out", "w", stdout);
    #endif

    cin >> n;
    for (int i = 1; i <= n / 2 + 1; ++i)
        cin >> x, q.push(x);
    for (int i = n / 2 + 2; i <= n; ++i){
        cin >> x;
        if (x < q.top()) q.pop(), q.push(x);
    }

    if (n % 2) printf ("%d.0", q.top());
    else{
        unsigned int top1 = q.top();
        q.pop();
        if ((top1 + q.top()) % 2 == 0) cout << top1 / 2 + q.top() / 2 << ".0";
        else cout << top1 / 2 + q.top() / 2 << ".5";
    }
    ioi
}

Edited by author 01.12.2016 20:51

Edited by author 01.12.2016 20:51

#include<stdio.h>
int main(void){
int a, b;
scanf("%d %d", &a, &b);

 if(a%2==0 || b%2!=0)
    puts("yes");
else if(a%2!=0 || b%2==0)
    puts("no");
return 0;
}