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);
}

If n = 1 answer is 0,  but accepted solution with answer 1.
Why???

"отделённых друг от друга ПРОИЗВОЛЬНЫМ КОЛ-ВОМ пробелов и переводов строк" - это меня "убивает"...

По сути ведь получается, что непонятно, когда заканчивается ввод исходных данных. Не так ли?

Или я могу чего-то не знать?
<python>

#include "stdafx.h"
#include<stdio.h>
#include<math.h>

int _tmain(int argc, _TCHAR* argv[])
{
    float a;
    scanf_s("%f", &a);
    printf_s("%6.4f", sqrt(a));
    scanf_s("%f", &a);
    return 0;
}

a=input().split()
t=0
for i in range(len(a)):
  c=a[t]
  v=int(c)
  a[t]=v
  t=t+1
e=[0]
s=[0]
z=[0]
q=[0]
y=1
for i in range(a[1]):
    d=int(input())
    s.append(d)
s.remove(s[0])
for i in range(0,a[0]):
    w=s.count(y)
    z.append(w)
    w=0
    y=y+1
z.remove(z[0])
t=0
for i in range(len(z)):
    c=z[t]
    v=int(c)
    z[t]=v
    t=t+1
t=0
c=0
for i in range(len(z)):
    c=z[t]/a[1]
    c=c*100
    q.append(c)
    t=t+1
q.remove(q[0])
t=0
c=0
for i in range (len(q)):
    c=round(q[t],2)
    e.append(c)
    t=t+1
e.remove(e[0])
t=0
for i in range (len(e)):
  print(e[t],"%")
  t=t+1

Edited by author 07.01.2018 14:22

Edited by author 07.01.2018 14:23