Edited by author 01.02.2020 21:31

using System;
using System.Linq;

namespace Csharp_1567
{
    class Program
    {
        static void Main(string[] args)
        {
            string str1 = Console.ReadLine();
            uint sum = 0;
            int lin = (from n in str1
                       select n).Count();
            for (int i =0;i<lin;i++)
            {
                if (str1[i] == 'a' || str1[i] == 'd' || str1[i] == 'g' || str1[i] == 'j' || str1[i] == 'm' || str1[i] == 'p' || str1[i] == 's' || str1[i] == 'v' || str1[i] == 'y' || str1[i] == '_' || str1[i] == '.' || str1[i] == ' ')
                {
                    sum += 1;
                    continue;
                }
                else if (str1[i] == 'b' || str1[i] == 'e' || str1[i] == 'h' || str1[i] == 'k' || str1[i] == 'n' || str1[i] == 'q' || str1[i] == 't' || str1[i] == 'w' || str1[i] == 'z' || str1[i] == ',')
                {
                    sum += 2;
                    continue;
                }
                else sum += 3;
            }
            Console.WriteLine(sum);
        }
    }
}

I got WA 1, but when I test this code myself, I always get a correct answer. What's wrong?

#include <iostream>
using namespace std;

int main()
{
    char speech[1001];
    cin.getline(speech, 1001);
    int cost = 0;
    for (int i = 0; i < 1001; i++)
    {
        switch (speech[i])
        {
        case 'a': case 'd': case 'g': case 'j': case 'm': case 'p': case 's': case 'v': case 'y': case '.': case ' ':
            cost += 1;
            break;
        case 'b': case 'e': case 'h': case 'k': case 'n': case 'q': case 't': case 'w': case 'z': case ',':
            cost += 2;
            break;
        case 'c': case 'f': case 'i': case 'l': case 'o': case 'r': case 'u': case 'x': case '!':
            cost += 3;
            break;

        }
    }
    cout << cost << endl;

    return 0;
}

Edited by author 29.08.2018 19:40

https://umath.ru/theory/priznak-delimosti-na-7/

s = 0
k = 1
x = input()
while len(x) > 3:
    s += int(x[-3:]) * k
    k *= -1
    x = x[:-3]
s += int(x) * k
print(s % 7)

Edited by author 25.01.2019 02:16

I didn't do anything for 'arbitary' sequence. But my solution got accepted.Can anybody help me clear this doubt?

Edited by author 22.01.2019 18:38

Edited by author 09.03.2021 22:26

10
6
1 2 even
1 1 even
3 4 odd
5 6 even
1 6 even
7 10 odd

i think the answer of this data should be 1,but my program answer is 4 but it return an accepted!

sort(a + 1, a + 1 + n);
  reverse(a + 1, a + 1 + n);

  for(int i = 1; i <= n; i++){
    if(sum1 > sum2){
      sum2 += a[i];
    }
    else{
      sum1 += a[i];
    }
  }

  cout << abs(sum1 - sum2);

If you write on c++ you can do this:

char dot;
unsigned int x,ip;

ip = 0;

for (int i = 24; i >= 8; i -= 8) {
  cin >> x >> dot;
  x <<= i;
  ip |= x;
}
cin >> x;
ip |= x;
------------------
Now you have ip, same you can read mask.

Edited by author 21.01.2019 14:50

Edited by author 21.01.2019 14:52

На Паскале не работает код если начать его так:

read(n); read(s); (или readln(s))

n - longint, s - string

Но если так:

readln(s);

А затем разрезать на n и s1, то всё ОК.
Разве так и должно быть?

#include <iostream>
using namespace std;
 long long  a,b,t,c,d,k,s,f;

int main ()
{
    c = 1; k = 0;
   cin >>t;
   for (t;t>=1;--t) {
     cin >>a;
      cin >>b;
   if ( b%a==0)
   {
       for( ; d != b; ++c )
       {
         d = a*c;
          if ( b % d == 0)
          {
              k= k+1;
              a = d;
              c = 1;
          }
       }
       cout << k<<"\n";
   }
   else
      cout << "0\n";
   c=1; k=0;
   }
    return 0;

}

Hi there!
I'm curious regarding submission metrics time and memory. Is this the avg time and avg memory of all the tests or maximum time and maximum memory used in the worst test?

Can anybody please reveal how does the system count time and memory for different languages? I would like to reproduce it locally before the submission. I use golang.
Thanks

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace phrase
{
    class Program
    {
        static int First(string input, string[]array)
        {
            int col = Convert.ToInt32(input.Length);
            if (col>0&&col<=1000)
            {
                int countArr = 0,countStr = 0,c = Convert.ToInt32(array.Length),count = 0,indexStr=0;
                for (; countArr < c; countArr++)
                {
                    for (; (indexStr = input.IndexOf(array[countArr], countStr)) !=-1; countStr = indexStr + 1)
                    {
                        count++;
                    }
                    countStr = 0;
                }
                return count;
            }
            return 0;
        }
        static void Main(string[] args)
        {
            string[] one = new string[] { "a", "d", "g", "j", "m", "p", "s", "v", "y", ".", " " };
            string[] two= new string[] { "b", "e", "h", "k", "n", "q", "t", "w", "z", "," };
            string[] three = new string[] { "c", "f", "i", "l", "o", "r", "u", "x", "!" };
            string input = Console.ReadLine();
            int a = First(input,one);
            int b = First(input,two);
            int c = First(input,three);
            Console.WriteLine(a+(b*2)+(c*3));
        }
    }
}

#include<stdio.h>
#include<stdlib.h>
#define x 100000

int main()
{
    long long int n , c=0,i;
    double ar[x] = {};

    while(scanf("%lld",&n)){
        if(n < 0) break;
        ar[c++] = n;
    }

    for(i=c-1; 0<=i; i--){
        printf("%.4lf\n",sqrt(ar[i]));
    }

    return 0;
}

var

  i, n: integer;
  c : real;

begin
  read(n);
  c := 0;
  if n > 0 then
  begin
    for i := 1 to n do
      c := n;
    write(c);
  end;
  if n <= 0 then
  begin
    for i := 1 downto n + 1  do
      c := (-((-n) * (1 - n) / 2) + 1);
    write(c);
  end;
end.

I think the answer is probablye Yes...

Be sure that(at least for the given interval, [4,2*10^9]) every even number can be shown as sum of two prime numbers.

1. if n is prime, {n}
2. if n is even, {n = p1 + p2}
3. if n is odd   {n = 3 + p1 + p2} ( n-3 is even, so that recall second case.)

For convenience, prime numbers under 10^5 can be initialized.

Edited by author 03.12.2012 03:00

Edited by author 03.12.2012 03:00