If you have troubles with test 4 than READ OUTPUT FORMAT CAREFULLY :)

What is the number 1 stand for?
and test # 2 ?
3 ?
4 ?

I was very lazy and made the program as slow as possible - using array instead of tree for storing numbers and Scanner for reading files and still got Accepted. How come?

We can write some FOCs which describe optimal point. But how to find point which satisfies these FOCs?

I send program with such procedure:

void moveNorth(double dist)
{
  w += dist / rEarth * 180 / pi;
  if (w > 91.0)
    n = (n - n) / n;
}

And got crash 15, but when i send

void moveNorth(double dist)
{
  w += dist / rEarth * 180 / pi;
}

i got accepted.

So, i think device can reach north pole with test 15 input data, but it's impossible.

who can help me?
I use bfs.

Please, give me test 12. Very need.

Don't forget about it ;)

If WA8, try test
100000
PUSH 1 1
PUSH 1 2
....
PUSH 1 99990
POP 1
...
POP 1
Right answer:
99990
...
99981

What answer for
3
1
0
3
0
?

0
0
0

or

0

?

give me test plz

import java.util.Scanner;

public class Main {

    public void solucion(int num) {

        int aux = 0;
        if (num <0) {
            for (int i = num; i <= 1; i++) {
                aux += i;
            }
        } else {
            for (int j = num; j >= 1; j--) {
                aux += j;
            }
        }
        System.out.println(aux);

    }

    public static void main(String[] args) {
        Main obj = new Main();
        Scanner leer = new Scanner(System.in);
        int numero = leer.nextInt();
        if (numero < 10000) {
            obj.solucion(numero);
        }


    }
}

#include <iostream.h>
int main(){
    int num;
    while (cin>>num){
          int answer=0;
          if(num>=1){
                     answer = num*(num+1)/2;
                     printf("%d\n",answer);
                     }
          else{
               answer = (2-num)*(1+num)/2;
               printf("%d\n",answer);
               }
          }
          return 0;
    }
where ?

what`s wrong?


var n,i,res,c,max:longint;
a:array[0..100000] of integer;
begin
repeat
max:=0;
Readln(n);
a[0]:=0;
a[1]:=1;
for i:=1 to n do
begin
a[2*i]:=a[i];
a[2*i+1]:=a[i]+a[i+1]
end;
for i:=1 to n do
if a[i]>max then max:=a[i];
if n<>0 then Writeln(max);
until n=0;
end.

Subject

This code gets WA on test 2:
[code deleted]

& this gets AC:
[code deleted]

:)

Edited by moderator 02.01.2020 18:26

I know O(n^2) one. Maybe there is also an O(1)?

#include <stdio.h>

int main(void) {
int n,k,i;
char c;
scanf("%d",&n);
while(n--){
scanf("%1s%d",&c,&k);
if((c=='a'||c=='h')&&(k==1||k==8)) {printf("2\n"); continue;}
if((c=='a'||c=='h')&&(k==2||k==7)) {printf("3\n"); continue;}
if((c=='b'||c=='g')&&(k==1||k==8)) {printf("3\n"); continue;}
if((c=='b'||c=='g')&&(k==2||k==7)) {printf("4\n"); continue;}
if(c<'b'||c>'g') {printf("4\n"); continue;}
if(k<2||k>7) {printf("4\n"); continue;}
if((c=='b'||c=='g') || (k==2||k==7)) {printf("6\n"); continue;}
printf("8\n");
}
return 0;
}

            long stringLength = Convert.ToInt32(Console.ReadLine());
            string S = Console.ReadLine();
            string T = Console.ReadLine();
            long temp = 0;
            bool check = false;
            for (long i = 0; i < stringLength; i++)
            {
                S = S[Convert.ToInt32(stringLength) - 1] +
                                     S.Substring(0,  Convert.ToInt32(stringLength) - 1);
                if (S == T)
                {
                    temp = i + 1;
                    check = true;
                    break;
                }
            }
            if (check && temp < stringLength)
                Console.WriteLine(temp);
            else
                Console.WriteLine(-1);

The printing of the answer in C++ should be, e.g., in the form:
cout << "Distance is "
     << setiosflags(std::ios_base::fixed)
     << setprecision(0)
     << 1000.0*sqrt(dist) << " km.";
where dist is double value. You don't need to round off the answer with the help of some functions.