i just draw its in paint and use some kinematick formulas and solve it by that

These two tests helped we find my bugs.

Test
2
1
3
0
2 1 2
The answer is 4.

Test
3
10
20
30
1
29 3 1 2 3
2 1 2
The answer is 29.

Hope it will help you too.

if you use the two-pointer method, then you have problems with the left pointer

N =int(input())
Nl = []

for i in range(N):
    Nl.append(int(input()))

M = int(input())
Ml = []

for j in range(M):
    Ml.append(int(input()))

answerl = []


Nl = set(Nl)
Nl = list(Nl)
Nl.sort()

for ni in Nl:
    for mi in Ml:
        if mi == ni:
            answerl.append(mi)
        else:
            continue

print(len(answerl))

#include <stdio.h>
    int main(){
        int a, b, c;
        scanf("%d", &a);
        b=12-a;
        if(b*45<=240){
            printf("Yes");
        }
        else{
            printf("No");
        }
        return 0;
    }

Why

3
3 2

n = int(input())
k=int(input())
n=n*2
if n % k ==0:
  print(n/k)
else:
    a=int(n/k)+1
    print(a)

I can't find my mistake.
It keep telling me "Runtime error"

Help with wa48

Why wa48 or wa49?

Edited by author 06.10.2017 11:52

program de;
    var
    a,b:array[-1400..1400] of integer;
    a1,b1:array[1..2800] of integer;
    i,j,n,max_a,max_b:integer;
    ch:char;
procedure diog;
begin
        if ch='s' then
            begin
             b[i-j]:=0;
             inc(a[i-j]);
             if a[i-j]>max_a then max_a:=a[i-j];
            end
         else
           begin
             a[i-j]:=0;
             inc(b[i-j]);
             if b[i-j]>max_b then max_b:=b[i-j];
            end;
end;
procedure diogobrat;
begin
        if ch='s' then
            begin
             b1[i+j]:=0;
             inc(a1[i+j]);
             if a1[i+j]>max_a then max_a:=a1[i+j];
            end
         else
           begin
             a1[i+j]:=0;
             inc(b1[i+j]);
             if b1[i+j]>max_b then max_b:=b1[i+j];
            end;
end;
procedure vivod;
begin
   if max_a=max_b then
      begin
        writeln('?');
        writeln(max_a);
      end
   else
   if max_a<max_b then
      begin
        writeln('S');
        writeln(max_b);
      end
   else
      begin
        writeln('s');
        writeln(max_a);
      end;
end;
begin
{   assign(input,'c:\test.txt');
   reset(input);                   }
   readln(n);
   for i:=1 to n do
   begin
     for j:=1 to n do
      begin
        read(ch);
        diog;
        diogobrat;
      end;
      readln;
   end;
   vivod;
end.

Please anyone help

try this test

3
2 4
1 5
3 5

bonnon
Answer
b onno n

Edited by author 27.10.2022 10:18

Edited by author 27.10.2022 10:19

n = int(input())
k=int(input())
n=n*2
if n % k ==0:
  print(n/k)
else:
    a=int(n/k)+1
    print(a)

I can't find my mistake.
It keep telling me "Runtime error"

My program got WA8 all the time.
I have added next string to my program

    if(n>50)         vector<long> xxx(1000000);

and it got AC after that!!!
Variable "xxx" never uses in my code! If i delete this line it got WA8!!!!

-200 -5 0
200 -5 0
0 200 0

answer should be 16707.55 (> pi*r*r/2), not 14708.38 (my AC program)

короче сам голову ломал, там переменные в 1 строку вводить надо
может кто то также ошибся

Edited by author 17.10.2022 09:55

Edited by author 17.10.2022 09:55

I don't know why there were so many submitted O(n^2) DP solutions. Here is a simple formula for this problem:
answer = n - (yes + yes*(yes-1) + no*(no-1))/n, where yes = s-2*n and no = n-yes are the total numbers of occurrences of corresponding strings in answer.txt.
You can easily derive this if you try to think of probability of the same verdict for two arbitrary consecutive tests.