why greedy with catching two mins dont work
pseydotests:
#1:
test

test, universal

base
#2:

universal, base

base, universal

#3:
universal

base

test

universal

Try the test
8
1 7 4 6 3 5 2 8
and
8
8 2 5 3 6 4 7 1
.
Both tests have solutions.

double

1000000000 1000000000 1000000000
0 0 0
answer:
It is a kind of magic

Please give me some tests, my program passed all the tests on the forum

Edited by author 22.07.2020 04:11

you don't know how to write bfs, try watching videos for beginners

precision

Try this test:
9 6
2 3 2 5 3 4 1 6 1

Answer:
8
1 6 1 2 2 5 3 4 3

try this:

sample input:
2
-1 -2
-3 -4

sample output:
-1

Код идеален но не проходил 5 тест. Писал на пайтоне. Просто ради эксперимента к sys.stdin.read() добавил стрип - sys.stdin.read().strip() и вуаля - AC. По условиям задачи, замечу, вроде так быть не должно. Если нужен сам рабочий код - пишите

I spent 20 submissions, because of TLE15. I tried many fft implementations. But the error was in my z function

In a task it is told that a way of departure (1; 1) though, actually, the way of departure has to be (0; 0)

My bee had to make only 200 flights, if you know what I mean :)
100 flights wasn't enough to get AC

n = int(input())
sp = []
c = 0
for i in range(n):
    b = input()
    sp.append(b)
sp.sort()
for i in sp:
    if sp.count(i) >= 2 and c == 0:
        print(i)
        c = 1
    if sp.count(i) < 2:
        c = 0

Pascal


var a,b: array[1..100] of string;
    i,j,k,n: integer;
    s: string;
    f: boolean;
begin
readln(n);
for i:=1 to n do
 begin
 readln(s);
 f:=false;
 for j:=1 to n do
  if a[j] = s then f:=true;
 if f = false then a[i]:=s
 else begin
      f:=false;
      for j:=1 to n div 2 do
       if b[j] = s then f:=true;
      if f = false then b[i]:=s;
      end;
 end;
for i:=1 to n div 2 do
 if b[i] <> '' then writeln(b[i]);

end.

please can anyone give me test #13?

no good epsilon for this problem
use big integers.

Using long long hashes, any ideas?

brute-force solution with bitsets gets AC in 0.187 sec

I've solved this problem finally, of course, but I still suppose its definition to be unclear a bit :) So some tests for you:

1 3
1 1 1

1

//

1 5
1 2 3 4 5

//

1 5
5 4 3 3 3

1

//

1 7
1 2 2 3 3 4 5

1

//

1 5
1 2 3 2 1

2

//

1 6
1 2 3 2 3 4

2 - that's the point!

//

1 6
3 2 1 4 4 5

2

//

1 6
1 2 1 2 1 2

3

//

1 6
1 2 3 1 2 1

3