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

use "long double"
and to avoid rounding when you print the result, use
cout << fixed <<setprecision(0)<< result << endl;

WA6?

Edited by author 21.01.2019 19:40

If it is a real tree, why after N lines, describing conviviality ratings of employees,
we have not simply (N - 1) lines, describing supervisor relations?

Does such input format mean that some employees may have more than one supervisors?

Edited by author 21.06.2012 18:08

Why do I memory limit exceeded even if I have use pointer?
Here is my code:

const mn=6000;
      inf=-100000000;
type rec=record
           father:integer;
           num:integer;
           child:array[1..mn]of ^integer;
         end;
var value:array[1..mn]of ^integer;
    f:array[1..mn,1..2]of ^longint;
    t:array[1..mn]of ^rec;
    n,i,a,b,root:integer;
    ans:longint;

function max(a,b:longint):longint;
  begin
    if a>b
      then exit(a)
      else exit(b);
  end;

function dp(root,lab:integer):longint;
  var i:integer;
  begin
    if root=0 then exit(0);
    if f[root,lab]^>inf then exit(f[root,lab]^);
    if lab=1
      then begin
        f[root,lab]^:=value[root]^;
        for i:=1 to t[root]^.num do
          inc(f[root,1]^,dp(t[root]^.child[i]^,2));
      end
      else begin
        f[root,lab]^:=0;
        for i:=1 to t[root]^.num do
          inc(f[root,2]^,max(dp(t[root]^.child[i]^,1),dp(t[root]^.child[i]^,2)));
      end;
    exit(f[root,lab]^);
  end;

begin
  readln(n);
  for i:=1 to n do
    begin
      new(value[i]);
      new(t[i]);
      new(f[i,1]);
      new(f[i,2]);
      readln(value[i]^);
    end;
  while not eof do
    begin
      readln(a,b);
      if(a=0)and(b=0)then break;
      t[a]^.father:=b;
      inc(t[b]^.num);
      new(t[b]^.child[t[b]^.num]);
      t[b]^.child[t[b]^.num]^:=a;
    end;
  for i:=1 to n do
    if t[i]^.father=0
      then begin
        root:=i;
        break;
      end;
  for i:=1 to n do
    begin
      f[i,1]^:=inf;
      f[i,2]^:=inf;
    end;
  ans:=max(dp(root,1),dp(root,2));
  writeln(ans);
end.

Who can HELP me?

Edited by author 05.04.2010 16:57

Some test cases that helped me:

6 4
001000 100001
000100 000010
010000 100100
100000 111000
100000
ans:
111111


6 4
010000 100000
001000 100001
100000 111000
000100 000001
100000
ans:
111001


6 4
100000 111000
010000 100000
001000 100001
000100 000001
000000
ans:
000000


6 4
111100 111000
111110 100000
000000 100001
111111 000001
010100
ans:
110101


6 4
111000 101000
111100 111000
111110 100000
111111 000001
011100
ans:
011100


6 0
010100
ans:
010100