#include <iostream>
#include <cmath>
using namespace std;
int fac(int n){
    if (n==0){
        return 1;
    }
    else {
        return n*fac(n-1);
    }
}
int wtf(int n,int k){
    return fac(n)/fac(n-k);
}
int cock(int n){
    int sum;
    for (int i=2;i<=n;i++){
        sum += wtf(n,i);
    }
    return sum;
}
int main()
{
    int n;
    cin >> n;
    cout << cock(n);
    return 0;
}

thanks

Hey,could you help me?

1 1 1
2 2 11
2 4 2
3 3 111
3 5 12
3 9 3
4 4 1111
4 6 112
4 8 22
4 10 13
4 16 4
5 5 11111
5 7 1112
5 9 122
5 11 113
5 13 23
5 17 14
5 25 5
6 6 111111
6 8 11112
6 10 1122
6 12 222
6 14 123
6 18 33
6 20 24
6 26 15
6 36 6

I have no ideas...
Please, give me some tests!

Try below test (TWO edges between the first and the last vertices):

2 2
1 2 60 60
1 2 50 51
1

Answer:
0.000000 1
1


Edited by author 21.12.2018 01:56

Edited by author 05.06.2020 17:42

I can solve it by breadth-first search with Execution time:0.218, but how to solve it with Execution time:0.015 or 0.031??? I have no idea about it. Give me please idea or algorithm's  name to solve it so fast. thank you!

Give tests, please!!

Прошу помочь.
Написал алгоритм на питоне, все придуманные мной примеры работают, однако шестой тест неумолим...
Подскажите, где ошибка? Код прилагаю

string = input()
find = 'Sandro'
lst = []
string = '      ' + string + '      '
def counter(indexInFind, indexInString):
    global string
    global find
    summ = 0
    index = 0
    temp=''
    for i in range(indexInString - indexInFind, indexInString - indexInFind + 6):
        if string[i] == ' ':
            return #
        if string[i] != find[index]:
            summ += 5
        index += 1
        temp += string[i]
    return summ if temp.find('S') != -1 or temp.find('s') != -1 else summ+5
for i in range(len(string)):
    strfind = find.find(string[i])
    if strfind != -1:
        lst.append(counter(strfind, i))
print(min(lst) if len(lst) != 0 else 35)

Edited by author 19.12.2018 01:29

Edited by author 19.12.2018 01:29

Edited by author 19.12.2018 01:31

a=int(input())
b=int(input())
if (a%2) == 0 or (b%2) == 1:
    print('yes')
else:
    print('no')

помогите найти ошибку. вроде ответы правильные выдаёт


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

namespace ConsoleApp3
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] a = Console.ReadLine().Split();
            int b = Convert.ToInt32(a[0]);
            int c = Convert.ToInt32(a[1]);
            int d = 0;
            string[] e = Console.ReadLine().Split();
            for (int f = 0; f < c; f++ )
            {
                d = d + Convert.ToInt32(e[f]);
            }
            int g = d - b * c;
            if (g < 0)
            {
                g = 0;
            }
            Console.WriteLine(g);
            Console.ReadLine();
        }
    }
}

Am I stupid or what?
code seems to be working

var x,a,b:integer;
function sum(x:integer): INTEGER;
  var s:integer;
  begin
  while x>0 do
    begin
    s:=s+x mod 10;
    x:=x div 10;
    end;
  sum:=s;
  end;
begin
read (x);
a:=x div 1000;
b:=x mod 1000;
if (sum(a)=sum(b+1)) or (sum(a)=sum(b-1)) then writeln ('Yes')
else writeln ('No');
end.

Can someone give me test #18??? Or give me some tests and answers???

I didn't find a mistake in my program. And what is the test #19? I'am confused(

tovars,prices,recomendations= [],[],[]
for i in range(6):
    name = input()
    tovar = input()
    price = int(input())
    if tovar in tovars:
        if prices[tovars.index(tovar)] > price:
            prices[tovars.index(tovar)] = price
        recomendations[tovars.index(tovar)] +=1
    else:
        tovars.append(tovar)
        prices.append(price)
        recomendations.append(1)
if recomendations.count(max(recomendations)) == 1:
    print(tovars[recomendations.index(max(recomendations))])
else:
    ans = []
    ans1 = []
    l = len(tovars)
    mm = max(recomendations)
    for i in range(l):
        if recomendations[i]  == mm:
            ans.append(tovars[i])
            ans1.append(prices[i])
    print(ans[ans1.index(min(ans1))])

Could you please add Common Lisp as a language? SBCL is the most popular, and most performant, implementation of Common Lisp, and has been for some time (forked from CMUCL nearly 20 years ago)! It is also mostly maintained by Stas Boukarev. It has static typing, and it would be a beautiful language to solve Timus problems in. Common Lisp has been around for a very long time and its ANSI Standard has been the same for quite a number of years, so it is possible that there would not need to be many updates after it is added once.

Adding the SBCL implementation as a supported language would be greatly appreciated by me and the Common Lisp community to be able to solve the amazing problems on this website
with this very performant meta programming language. While attempting to get SBCL added, this repo could be used for reference -- https://github.com/optimisticlisper

I really hope you will consider it.

#include<stdio.h>
#include<math.h>
int main()
{
    int  n;

    printf("%d\n",scanf("%d",&n)*n*scanf("%d",&n)*n*scanf("%d",&n)*n*2);
}

At first,I got wa on test1,like many other authors.But later,I found it!
All in one sentence,test the data below:


1
inputon

Create struct:

struct Animal which has:
- ID // ID of hero
- name // name of animal (char *) (cin.getline) (cin.ignore())
- strength(int) // power of animal
- frequency // number of hits in one iteration
- HP // amount of life

Your task is to show animal which won most rounds in the game.

Input:
n (how many animals) [2 <= n <= 10]
n-times:
ID
name
strength
frequency goals
HP goals
Output:
winner with amount of victory.

Example:

Input:
10
1 Lion John 30 2 100 // (0 1 0 0 1 0 1 0 0) 3 victory
2 Hen Rebecca 60 1 200 // (1 1 1 0 1 0 1 0 1) 6 victory
3 Tiger Alex 40 2 100 // (1 0 0 0 1 0 1 0 1) 4 victory
4 Buffalo James 40 3 100 // (1 1 1 0 1 0 1 0 1) 6 victory
5 Elephant Jack 50 1 300 // (1 1 1 1 1 0 1 0 1) 7 victory
6 Rhino Michael 30 1 200 // (1 0 0 0 0 0 1 0 0) 2 victory
7 Snake Oz 50 2 400 // (1 1 1 1 1 1 1 1 1) 9 victory
8 Cow Bettie 40 1 100 // (0 0 0 0 0 0 0 0 0) 0 victory
9 Crocodile Nick 20 3 500 // (1 1 1 1 1 1 0 1 1) 8 victory
10 Bear Drake 40 2 150 // (1 1 1 1 0 1 0 1 0) 6 victory

Output :
7 Snake Oz 50 2 400 9 victory