I'd like to thank authors for this problem - it became one of my favorite on Timus!
Thanks again for very interesting problem!
:-)

Unfortunately entirely deterministic version of this algorithm fails on Test #10. So I'm very curious about this Test #10. Because I've done a lot of testing in MATLAB of varoius modification of Floyd-Steinberg dithering algo, including zig-zag proseccing and edge attending. In all cases constraint in statement can be hardened from 20 to 10 and even less. On normal images error is about 6-8.

I guess Test #10 some kind of periodic image, that has bad compatibility with that unnatural restriction put in the problem. I mean it square form and discontinuous nature. I think dithering yield much more better image than that which can be produced by uniform minimization of constrain function.

I solve this issue whit Test #10 by adding a little bit of randomness in my dithering algo. I mean in place of this threashold function:

int quantize(int value) {
    if (127 < value) {
        return 255;
    }
    return 0;
}

I'm using this function:

int quantize(int value, int randomness) {
    if (127 - randomness + rng.nextInt(2 * randomness + 1) < value) {
        return 255;
    }
    return 0;
}

This provides me with magical parameter "randomness" varying which and playing tambourine I got my AC. 8-] I think I should add constraints checking and reprocessing in case the check was a fail, but I'm too lazy, I guess in case of possible future rejudging everything will be fine. :)

I express my sincerest thanks to authors of this beautiful problem. It has bugged me a lot in era of matrix printers how can they do this funny dotted images. Finally I have my answer and I can even do it myself!!!

Edited by author 17.01.2018 02:43

#include <iostream>
using namespace std;

int main()
{
    int k, n, res=0;
    cin >> k >> n;
    int *mas = new int[n];

    for (int i=0; i!=n; i++)
    {
        cin >> mas[i];

        if (mas[i] > k)
        {
            res = res + (mas[i] - k);
        }
        else
        {
            if (res!=0 )
            {
                if ((k - mas[i]) > res)
                {
                    res = 0;
                }
                else
                {
                    res = res - (k - mas[i]);
                }
            }
        }

    }
    cout << res;

    return 0;
}

ll n, m, ans;
st s;
int main(){
  while(cin >> s){
    for(int i = 0;i < s.sz;++i){
      if(s[i] == 'b' || s[i] == 'e' || s[i] == 'h' || s[i] == 'k' || s[i] == 'n' || s[i] == 'q' || s[i] == 't' || s[i] == 'w' || s[i] == 'z' || s[i] ==',')
          ans+=2;
         if(s[i] == 'a' || s[i] == 'd' || s[i] == 'g' || s[i] == 'j' || s[i] == 'm' || s[i] == 'p' || s[i] == 's' || s[i] =='v'|| s[i] == 'y' || s[i] == '.')
             ans+=1;
         if(s[i] == 'c' || s[i] == 'f' || s[i] == 'i' || s[i] == 'l' || s[i] == 'o' || s[i] == 'r' || s[i] == 'u' || s[i] == 'x'|| s[i] == '!')
              ans+=3;
      }
      ans++;
      }
  cout << ans - 1;
    }

Why did you got wa on test 3 ? I'm having the same problem... :(

Только начал изучать С++. Решаю задачки для новичков. Что не так здесь?

#include <iostream>
#include <conio.h>

using namespace std;

int main()
{
    int A;

cout << "How many?";

cin >> A;

if(A < 1 || A > 2000) {
cout << "wrong input!";
}else { if(A <= 4)
        cout << "few";
else if(A <= 9)
        cout << "several";
else  if(A <= 19)
        cout << "pack";
else  if(A <= 49)
        cout << "lots";
else  if(A <= 99)
        cout << "horde";
else  if(A <= 249)
        cout << "throng";
else  if(A <= 499)
        cout << "swarm";
else  if(A <= 999)
        cout << "zounds";
else cout << "legion";}

return 0;
}

Edited by author 15.01.2018 06:34

WA 46:
2
0 0
0 0.5
Ans:
0 2

WA 47:
2
0 0
0 99
Ans:
1 1

WA 48:
2
0 0
0 1
Ans:
0 2
1 1

What is test case 12?

You need to compute two scalar products, each for 0 bit and 1 bit. Then just compare it. That'a ALL!!! If you are afraid of amplitude being negative then abs() each product and compare them after. There is no need to fuss about anything: carriers and constant level orthogonal to each other and noise guaranteed to be small by problem statement. Well, if noise was too big betrayer would never be able to transmit his data. Even if every bit was transmitted with differen amplitude and different constant level this approach still would work.

I just do not undestand why this problem is rated soooooo high?!

I really stuck with this, I don't know how to make my algorithm faster

here is my code:
#include <stdio.h>
char board[10][10];
char letter[66],num[66];
int n,s=0,ncuad;
int pos_f[8]={-1,-2,-2,-1, 1, 2, 2, 1};
int pos_c[8]={-2, 1,-1, 2, 2, 1,-1,-2};

init_board(){
    int i,j;
    for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
            board[i][j]='0';
}

search(int i,int f,int c){
    int j=0;
    if(s==1)
        return 0;
  if(board[f][c]=='0'){
     board[f][c]='1';
     switch(f){
           case 1: letter[i]='a';break;
        case 2: letter[i]='b';break;
        case 3: letter[i]='c';break;
        case 4: letter[i]='d';break;
               case 5: letter[i]='e';break;
               case 6: letter[i]='f';break;
               case 7: letter[i]='g';break;
               case 8: letter[i]='h';break;
     }
     num[i]=c;
     if(i==ncuad)
              s=1;
       while(j<8 && s==0){
           if((f+pos_f[j])>0 && (f+pos_f[j])<=n)
         if((c+pos_c[j])>0 && (c+pos_c[j])<=n)
               search(i+1,f+pos_f[j],c+pos_c[j]);
       j++;
     }
         board[f][c]='0';
   }
}

int main(){
    int i,j;
  scanf("%d",&n);
  ncuad=n*n;
  init_board();
     for(i=1;i<=n;i++)
         for(j=1;j<=n;j++)
           search(1,i,j);
    if(s){
       for(i=1;i<=ncuad;i++)
         printf("%c%d ",letter[i],num[i]);
         return 0;
    }
    else
     printf("IMPOSSIBLE\n");
  return 0;
}

if someone can help me I would appreciate a lot!!

The answer is correct, but it returns an error. Why? Maybe I did not understand the condition of the task correctly, or can the data be specified for which this code will produce the wrong solution?


using System;

namespace ConsoleApp30
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            int x = 0;
            string[] shoplist = new string[n];
            for (int i = 0; i <n; i++)
            {
                shoplist[i] = Console.ReadLine();
            }

            for (int i = 0; i < shoplist.Length; i++)
            {
                for (int j = i + 1; j<shoplist.Length; j++)
                {
                    if (shoplist[i] == shoplist[j])
                    {
                        x++;
                        i++;
                    }
                    else
                    {
                        continue;
                    }
                }
            }

            Console.WriteLine(x);
            //Console.ReadKey();
        }
    }
}

using System;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = Convert.ToInt32(Console.ReadLine());
            string [] ch= new string [100];
            string c;
            for (int i=0;i<n;i++)
                ch[i]=Console.ReadLine();
            c=Console.ReadLine();
            for (int i=0;i<n;i++)
            {
                if (c==ch[i].Substring(0,1)) Console.WriteLine(ch[i]);
            }
            Console.ReadKey();
        }
    }
}

what's the problem??? help me please

I mean not just to see on examples and find a pattern

Let's define C(k,n) = n!/(k!(n-k)!).

Binomial coefficients are widely used in combinatorics. The number of ways you can place something on something is an binomial coefficient. But we can't place the blue stripe on the end of the flag and side by side.

It is well-known, that
   C(0,n) + C(1,n) + ... + C([n/2],[n/2]) = F[n+1]
where F[n+1] is (n+1)-th Fibonacci number and [n/2] is integer division [1].

I hope this my small review does not spoil you the solving of the problem. Thank you.

[1] https://en.wikipedia.org/wiki/Fibonacci_number (Use in Mathematics)

accepted/sended = 13616/35461

Edited by author 11.01.2018 10:21

Edited by author 17.06.2019 01:04

Edited by author 17.06.2019 01:04

#include<stdio.h>
#include<math.h>
#define size 256*1024/sizeof(long long)
int main()
{
  int s=0,l;
  long long int a[size+1];
  do{l=scanf("%lld",a[s++]);}while(l!=EOF);
  for(s=s-2;s>=0;s--)
     printf("%.4Lf\n", sqrt((double)a[s]));

}

Try n = 1

// Harry Poter
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <string.h>
using namespace std;
int main()
{
    char line[200001], w2[200001];
    cin.getline(line, sizeof(line));
    int t = 0, i;

    for(i = 0; i < strlen(line); i++){
            if (line[i] == line[i + 1]) {
                i++;
A:    if (line[i + 1] != line[i + 2]) {
        if (t != 0 && w2[t - 1] == line[i + 1]) {t--; i += 1; goto A;}
                }
            }
    else {w2[t++] = line[i];}
        }
    w2[t] = '\0';
    printf("%s", w2);
}