I really appreciate your help guys, please give a test, I spend so much time to solve this problem, now I don't wanna leave it like that.

1)Pls, give me some tests
2)Am I right that answer < 10^18?

Edited by author 22.07.2019 23:42

def bin_Search(arr, val):
    low = 0
    high = len(arr)-1
    while low <= high:
        mid = (low + high) // 2
        if val < arr[mid]:
            high = mid - 1
        elif val > arr[mid]:
            low = mid + 1
        else:
            return True
    else:
           return False


first = ["Alice", "Ariel", "Aurora", "Phil", "Peter", "Olaf", "Phoebus", "Ralph", "Robin"]
first.sort()
second = ["Bambi", "Belle", "Bolt", "Mulan", "Mowgli", "Mickey", "Silver", "Simba", "Stitch"]
second.sort()
third = ["Dumbo", "Genit", "Jiminy", "Kuzko", "Kida", "Kenai", "Tarzan", "Tiana", "Winnie"]
third.sort()
n = int(input())
pos = 1
steps = 0
for i in range(n):
    name = str(input())
    if bin_Search(first, name) == True:
        steps += pos - 1
        pos = 1
    elif bin_Search(second, name) == True:
        if (pos - 2) < 0:
            steps += ((pos - 2) * -1)
            pos = 2
        else:
            steps += pos - 2
            pos = 2
    else:
        if (pos - 3) < 0:
            steps += ((pos - 3) * -1)
            pos = 3
        else:
            steps += pos - 3
            pos = 3
print(steps)


In this solution i used binary search

m, n = map(int, input().split())

if (m*n) % 2 == 0:
    print("[:=[first]")
else:
    print("[second]=:]")

import sys
from math import sqrt
potok = sys.stdin.read()
spisok = potok.split()[::-1]
if sys.getsizeof(potok)>262144:
    print('большой объем')
    quit()
for i in spisok:
    i=int(i)
    if i>10**18 or i<0:
        print('неверный диапозон')
        quit()
    res = sqrt(int(i))
    print("%.4f" % res)

Input
3 3 1
1 1
3 3
Output
0
is it correct?

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char* argv[])
{
    int co1=0,co2=0,co3=0,c=0; //init
    //1
    cin >> co1;
    int *nm1 = new int[co1];
    for (int i = 0; i < co1; i++)
    {
        cin >> nm1[i];
    }
    //2
    cin >> co2;
    int *nm2 = new int[co2];
    for (int i = 0; i < co2; i++)
    {
        cin >> nm2[i];
    }
    //3
    cin >> co3;
    int *nm3 = new int[co3];
    for (int i = 0; i < co3; i++)
    {
        cin >> nm3[i];
    }

    for(int i=0;i<co1;i++)
    {
        for(int k=0;k<co2;k++)
        {
          if(nm1[i]==nm2[k])
          {
              for(int j=0;j<co3;j++)
              {
                  if(nm2[k]==nm3[j])
                  {
                      c++;
                      break;
                  }
              }
              break;
          }
        }
    }
    cout << c;
    system("pause");
    return 0;
}

f = int(input())
friends = []
cnt  = 0
for i in range(f):
    j = str(input())
    if j[-4:] == "+one":
        cnt += 200
    else:
        cnt+= 100
if ((cnt/100)+2) == 13:
    cnt += 300
else:
    cnt += 200

print(cnt)

I solve it by this

After going through all the discussion topics, I found that this problem can be solved using different techniques. I have made a list of keywords, and I would like to know exactly how many different ways are there to solve this problem?

1) Brute force
2) Bit mask
3) Dynamic Programming
4) Backtracking
5) Balanced Partition
6) Partition problem
7) Greedy algorithm

My idea:
1) if triangle is right, answer is obvious)
2) if triangle is equilateral, there is no solution
3) in other cases try to split the triangle into 2 right triangles to get coordinates
Correct me if I'm wrong or give test which fail that algo.

program text;

{$APPTYPE CONSOLE}

var

  n,k,i,j,n1,n2,col,lastlen:integer;
  num1,num2:array[0..1000001] of byte;
  t1,t2:longword;

Procedure Add;
var
  i,carry:integer;
begin
  carry:=0;
  for i:=1 to high(num1) do
  begin
    carry:=carry+num1[i]+num2[i];
    if carry>=10 then
    begin
      num1[i]:=carry-10;
      carry:=1;
    end else
    begin
      num1[i]:=carry;
      carry:=0;
    end;
  end;
end;



begin
  { TODO -oUser -cConsole Main : Insert code here }
{ Assign(input,'Input.txt');
 reset(input);
 Assign(output,'output.txt');
 rewrite(output);

} fillchar(num1,sizeof(num1),0);
  fillchar(num2,sizeof(num2),0);

  readln(n);
  col:=0;
  k:=0;
  for i:=n downto 1 do
  begin
    readln(n1,n2);
    num1[i]:=n1;
    num2[i]:=n2;
  end;


    ADD;

  k:=high(num1);
  while num1[k]=0 do dec(k);

  if k=0 then writeln(0) else
  for i:=k downto 1 do  write(num1[i]);

end.

Edited by author 30.07.2019 19:29

#include <iostream>
#include <vector>

using namespace std;

int main()
{
    int n;
    cin >> n;

    int a;
    int b;
    long double c=0;
    for (unsigned int i = 0; i < n; i++)
    {
        cin >> a;
        cin >> b;

        if (i != 0 )
        {
            c = c * 10 + a + b;
        }
        else
            c = a + b;


    }


    cout <<c<< endl;


}

What is correct algorithm?

My program that find 2 mins and print Math.Max(min1 + min2 - n, 0) don't pass test 13.

How did the thief steal the bicycle ?I'm not pretty sure.   Is there some one telling me ?

Sorry for my bad English.
I use

    int ch;
    while ((ch = System.in.read()) != -1) {
        // my solution
        System.out.print(<some char>);
    }
    System.out.println();

But I get WA #1 with or without last code line. I also used System.out.flush (without last line), but it do not help me.

Edited by author 24.06.2019 13:55

hEdited by author 19.07.2019 12:15


Edited by author 19.07.2019 12:16

import math

nm = input().split()
n = int(nm[0])
m = int(nm[1])
res = set()
result = []
for i in range(m):
    x = int(input())
    if x not in res:
        res.add(x)
    result.append(x)
answer = []
for i in res:
    answer.append(result.count(i))
while n - len(answer) > 0:
    answer.append(0)
for i in answer:
    v = str(round(i*100/m, 2))
    if v[-2] + v[-1] == '.0':
        print(str(round(i*100/m, 2)) + '0%')
    else:
        print(str(round(i*100/m, 2)) + '%')














Edited by author 24.02.2019 01:39

Edited by author 09.03.2019 03:19