Bruteforcing length gives WA, bruteforcing angle gives AC

I checked several times and everything is working correctly, help me


def encryption(txt):
    cipher = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}


    encryption_list = []
    for i in txt:
        encryption_list.append(cipher[i])

    c = d = c1 = 0
    encryption_last_list = []
    for i in encryption_list:
        if d == 0:
            d += 1
            c = i
            i -= 5

            if i < 0:
                i = 25 - abs(i)
            encryption_last_list.append(i)

        else:
            c1 = i
            i -= c
            c = c1
            if i < 0:
                i = 26 - abs(i)
            encryption_last_list.append(i)

    decodding_list = []
    for i in encryption_last_list:
        for k, v in cipher.items():
            if v == i:
                decodding_list.append(k)

    return decodding_list

text = input()
if len(text) <= 99:
    c = ''
    output = encryption(text.lower())
    for i in output:
        c += i
    print(c)

can anyone please explain the solution which uses the coordinates i think ...

Edited by author 24.01.2023 20:46

Edited by author 24.01.2023 20:46

Edited by author 24.01.2023 20:46

I get WA 1 with calculating distance in this way
for(int i=0;i<len;i++)
            {
                ans+=pow(((int)(buf[i]-'0')),3);
            }
while accepted in this way
 for(int i=0;i<len;i++)
            {
                ans+=(buf[i]-'0')*(buf[i]-'0')*(buf[i]-'0');
            }
WHY?!!!!
Is this a bug on judge or bug in function?!!
the compiler is g++

Notice, that +1 and -1 mean not victory and defeat states.

INPUT
2
L 1 +1

OUTPUT
+1

4 3
1 2
2 3
3 4

Luck is possible

be attentive with loop (while)! we know formule n - x * k =) and in depends of number parity condition will be >= (ex. [n - x * k >= n mod k] or [n - x * k >= k]). Good luck!)

Edited by author 19.01.2023 16:23

Can anyone check if I missed any scenario ?

1, 2, 3, 4, 5 => Nothing to do here"
5, 4, 3, 2, 1 => Nothing to do here"
5, 2, 3, 4, 1 => "Yes\n1 5"
1, 8, 3, 4, 5, 6, 7, 2, 9 => "Yes\n2 8"
1, 8, 7, 6, 5, 4, 3, 2, 9 => "Yes\n1 9"
9, 8, 3, 6, 5, 4, 7, 2, 1 => "Yes\n3 7"
1 => Nothing to do here"
1, 1, 1, 1, 1 => "Nothing to do here"
9, 2, 3, 5, 4, 6, 7, 8, 1 => "No hope"
1, 2, 3, 9, 5, 6, 7 => "No hope"
7, 6, 5, 9, 3, 2, 1 => "No hope"
1, 1 => "Nothing to do here"
1, 2 => "Nothing to do here"
1, 2, 4, 3, 5, 6 => "Yes\n3 4"
6, 5, 3, 4, 2, 1 => "Yes\n3 4"
2, 1, 3 => "Yes\n1 2"
1, 3, 2 => "Yes\n2 3"
1, 2, 3, 2, 3, 4, 5 => "Yes\n3 4"
1, 5, 7, 2, 4, 5 => "No hope"

WA-3:
1 2 3
.#
#.
..
1 1 1
1 1 3
-----
NO

WA-12:
3 3 3
.#.
...
...
..#
...
..#
...
...
...
1 2 2
3 3 3
-----
YES
2
climb east
jump south

Interesting that test48, test70 (with reservations) and test80 do not look so hard to my solution. At least if the commenters in previous threads were right with examples of these tests. But mysterious test30 and test31 - I don't even imagine what structure they got.

Test70 is "randomly" solvable: with time distributed between 0.3 and 2.9 seconds (with avg = 1.1 sec) on my Intel i5-9400 (by "randomly" - I mean, of course, that could be 2.9 seconds which is TLE and would not pass)

And my suggestion: It would be nice to have a separate section on the site with parallel running of tests for such hard computational tasks. I mean: maybe someone also passes test 70, but fails test 30, for example. And the total percentage: passed N tests, failed tests with numbers K, L, M

In this task you need to divide by 9. I wrote blog about this: https://codeforces.com/blog/andreyDagger

..................................................
.......................┌──────┐...................
.......................│......│...................
.......................│......│...................
.......................│......│...................
.......................│......│...................
.......................│......│...................
┌────────┐.............│......│...................
│........│.............└──────┘...................
│........│........................................
│........│........................................
│─────┐..│........................................
│.....│..│........................................
│.....│..│........................................
│.....│..│..............┌──┐......................
│.....│..│..............│..│......................
└────────┘..............│..│......................
└─────┘.................└──┘......................
..................................................
..................................................



.................................┌┐...............
.................................└┘...............
..┌──────┐.................................┌─────┐
..│......│........┌────────┐...............│.....│
..│......│........│........│...............│.....│
..│......│........│........│...┌───┐.......│.....│
..│.┌───┐│........│........│...│...│.......│.....│
..│.│┌──││───┐....│........│..┌│──┐│.......│.....│
..│.││..││...│....│........│..││..││.......└─────┘
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.│........│...└─────┘........│..┌│................
.└────────┘..................│..└│................
.............................└───┘................



..................................................
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..........┌─┐─┌─┐.................................
..........│.│.│.│.................................
..........┌─────┐.................................
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..........│.....│.....................┌┌┐.........
..........│.....│.....................┌┐┐.........
..........│.....│.....................└┘┘.........
..........└─────┘.................................
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..................................................
..┌┐..............................................
..└┌┐┌┐...........................................
...└┌└┘...........................................
....└┘............................................
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...............................┌┌┌┐┐┐.............
...............................└└└┘┘┘.............
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.....│┘─└│........................................
.....││.││........................................
.....│┐─┌│........................................
.....└───┘........................................
..................................................
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┌┌──┌┌──┐────┐┐───┌┐─┌───────┐─┌─────┌──┐┐───┌───┐
┌│──│││││─│──│──┌─────────┐───┐┌────┐─┐┌││┐││││┐┐│
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││┌┌│└──┘─│─┌│┐─││││││││││││┌─│││││─││││┘│││─││─││
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┌│││└──┘┌─└│││──└────┘││└┘│││││┌││└─│─│┘│─││────┘│
┌│┐││─┐│││││││││││││┌─┐│──││─│││││─┐││││││││││││││
│││─└──────││┘│┌│││││─││─┌││└│└─││──│─┘│││││─┐│┘││
└└─────││││││─│└││││└─┘│──│└────│┘││││─││┘│└─────┘
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││└│└───└──└──└┘─│─└──││└┘──││└│┘│││││─┘│└────┘│││
│└└│─│┘─│││││──││└────┘│─┘─││└─└───└────│──┘┌└─│─┘
└──┘─│└─││││└───│──│──┘│────│─┘│─└─┘──┘─└──────┘─┘
└────└────┘└────└──────┘────└──────┘┘┘───└┘─┘─┘┘└┘

Edited by author 08.09.2019 16:59

Edited by author 08.09.2019 17:01

Edited by author 08.09.2019 17:01

Edited by author 08.09.2019 17:06

Edited by author 08.09.2019 17:06

Edited by author 08.09.2019 17:07

Edited by author 08.09.2019 17:07

Edited by author 08.09.2019 17:15

#include <stdio.h>

#define maxn 1000001
        long n;
       short result[31],b[maxn];
       int tr[maxn];
      long count,i,j;

main()
{
       scanf("%ld",&n);
      for (i=0;i<n;i++) b[i]=0;
      b[1%n]=1;
      tr[1%n]=0;
      if (n!=1) b[2%n]=2;
      tr[2%n]=0;
      count=1;
       while (b[0]==0 && count <30)
      {
          count++;
          for (i=1;i<n;i++)
         if  (b[i]!=0)
         {
             if (b[(i*10+1)%n]==0)
            {
                b[(i*10+1)%n]=1;
               tr[(i*10+1)%n]=i;
            };
             if (b[(i*10+2)%n]==0)
            {
                b[(i*10+2)%n]=2;
               tr[(i*10+2)%n]=i;
            };
         }
      }
      if (b[0]==0)
      {
          printf("Impossible");
      }
      else
      {
          i=0;
         j=1;
         result[j]=b[i];
         while (tr[i]!=0)
         {
             i=tr[i];
            j++;
            result[j]=b[i];
         }
         for (i=j;i>0;i--) printf("%d",result[i]);
      }
    getchar();
}

var
  f,q,be:array[0..1000] of integer;
  n,x,a,b,c:integer;
  m:char;
begin
{assign(input,'1282.txt');
reset(input);
}  readln(n);
  for x:=1 to n do
    be[x]:=-3;
  for x:=2 to n do
    begin
      read(m);
      if m='L' then
        begin
          readln(a,b);
          if be[a]=-3 then
            begin
              inc(c);
              q[c]:=a;
              be[a]:=-1;
            end;
          if b=-1 then be[a]:=1;
          if (b=0) and (be[a]<0) then be[a]:=0;
        end
      else
        readln(f[x]);
    end;
  b:=1;
  while q[b]<>1 do
    begin
      if be[f[q[b]]]<-2 then
        begin
          inc(c);
          q[c]:=f[q[b]];
          be[f[q[b]]]:=-1;
        end;
      if be[q[b]]=-1 then be[f[q[b]]]:=1;
      if (be[q[b]]=0) and (be[f[q[b]]]<0) then be[f[q[b]]]:=0;
      inc(b);
    end;
  if be[1]=1 then writeln('+1') else writeln(be[1]);
end.
{be[n]==the best value it then player play best
f[n]==the father of n
be[n]=-3===>it hasn't been visited
}

The test 1 is different, and checks this condition.

I tried to figure it out, how 1 tree and two sticks with the length of 2 can form the area more than 4? The largest area would be rectangle. What is the trick on this problem, can anybody explain with Geometry language.

Thanks.

Test:
10
2 3 2 3 6
0 1 1
0 2 4 10
0 3 5 7 9
0 1 3
0 1 3
0 1 8
0 1 7
1 1 7
2 1 5

Answer:
unknown 2
2 2
2 2
2 2
2 2
2 2
unknown unknown
unknown unknown
2 1
2 2

Definition: "inside the island"

Let's define the definition recursively:
1)If an island/sea touches the edge of the map, then it does not lie inside the sea/island

2)If an island/sea does not have internal seas/islands, then all the seas/islands it touches (it should be one if it does not touch the edge of the map) contain this island/sea inside themselves.

3)If an island/sea has internal seas/islands, then all these seas/islands in this island/sea, and all the seas of which these island/sea touch, which are not internal, contain this island/sea.

this all means that if the island lies inside the sea, which is inside the island, which is inside the sea, which is inside the island that touches the end of the map, and the illuminated sea is selected, the sea that is inside the island touching the map, then all the islands inside the sea are not part of the island touching the end of the map, and if the outer sea is selected, then since it is not exactly contains an island touching the map, then it should not contain islands inside this island

If you get WA 30 check this:

1) correct calculation of tangent line angle (asin(2R/D), NOT asin(R/D))
2) check not only enemies circles to make your movement, but angles to your own circles too. You can destroy your circles and win by this angle (don't forget about this).

These tests helped me:

Test 1:
7.100 2.280 3.530 4.410 2.690 2.990 1.770 4.360 5.850 1.770 1.960 1.720 6.550 1.870 2.200 4.940
4.090 0.920 6.580 3.590 1.970 1.070 2.390 2.700 0.360 4.030 1.520 6.730 5.820 5.280 1.640 5.520
Answer: WHITE

Test 2:
2.670 6.880 2.140 2.820 0.830 6.960 2.180 4.600 6.220 7.170 7.130 1.290 3.010 6.320 2.880 4.690
0.350 6.670 6.680 5.820 7.230 5.020 7.160 4.100 0.120 0.430 2.950 7.780 5.630 2.400 2.290 1.310
Answer: RED

Edited by author 06.01.2023 21:45

just read some titles abot hanoi tower and learn some algorithm like floyd and u 100% will can solve this problem