use +E -E

This can be easily done on python

5 4
2 1 1
3 2 1
4 3 1
4 0 1

Impossible after 4 statements

This solution is OK. But it not satisfy the restrictions.
Your tests are too weak.

#include <iostream>
int main(int argc, char* argv[])
{
    int i= 0;
    for (i = 0; i < 1000000; i++)
            {
                printf("%c", rand()%26 + 'a');
            }
    return 0;
}

U HAVE VERY VERY VERY STUPID ERROR IN YOUR CODE

#include <bits/stdc++.h>
using namespace std;

int main(){
    long long n, res;
    vector <long long> vt;
    while(cin>>n){
        vt.push_back(n);
    }
    int len = vt.size();

    for(int i = len-1; i >= 0; i--){
        double n = sqrt(vt[i]);
        printf("%0.4lf\n", n);
    }

return 0;
}

Edited by author 07.09.2019 11:05

Edited by author 07.09.2019 11:11

   you can only use dfs , but don't forget to sort the team . When I
use sorting as from short to long , I got TL , but when from long to
short , only 0.02s !

Apparently, this is a very complicated problem,
but 1115 allows solutions with naive implementations.
I almost used brute force here, with some little optimizations (like looking for solutions for the next few sums while working on the 'current' one).

a=[int(i) for i in input().split()]
for i in a[::-1]:
    print(f"{float(i)**0.5:.4f}")

Some hints please. My algo takes too much time.

PLEASE TELL ME WHAT IS WRONG! I TRIED EVERYTHING!!!

#include<stdio.h>
#include<queue>
using namespace std;
struct Coord {int x,y;};
int a[40][40];
inline int number(int x,int y)
{
    int nr=0;
    if(a[x+1][y]==-1) nr++;
    if(a[x-1][y]==-1) nr++;
    if(a[x][y+1]==-1) nr++;
    if(a[x][y-1]==-1) nr++;
    return nr;
}
Coord make(int a,int b)
{
    Coord x;
    x.x=a;x.y=b;
    return x;
}
void lee(Coord s)
{
    int x,y;
    queue<Coord> q;
    q.push(s);
    a[s.x][s.y]=1;
    while(!q.empty())
    {
        x=q.front().x;y=q.front().y;
        if(a[x+1][y]==0) {a[x+1][y]=1;q.push(make(x+1,y));}
        if(a[x-1][y]==0) {a[x-1][y]=1;q.push(make(x-1,y));}
        if(a[x][y+1]==0) {a[x][y+1]=1;q.push(make(x,y+1));}
        if(a[x][y-1]==0) {a[x][y-1]=1;q.push(make(x,y-1));}
        q.pop();
    }
}
int main()
{
    int n,i,j;char ch;
    scanf("%d\n",&n);
    for(i=1;i<=n;i++,scanf("%c",&ch))
        for(j=1;j<=n;j++)
        {
            scanf("%c",&ch);
            if(ch=='.') a[i][j]= 0;
            else        a[i][j]=-1;
        }
    for(i=2;i<=n+1;i++)
        a[i][0]=a[0][i]=-1;
    for(i=0;i<n;i++)
        a[i][n+1]=a[n+1][i]=-1;
    lee(make(1,1));
    lee(make(n,n));
    int nr=0;
    for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
            if(a[i][j]==1)
                nr+=number(i,j);
    printf("%d\n",9*nr);
    return 0;
}

Or at least add another version of the problem with huge labyrinths.
I had a lot of fun with this problem, and was surprised that a solution which is far from perfect can work instantly with input 200x200.

3
.##
###
##.

In C#, I was getting TLE using LinkedList, so I rewrote everything using an array of 2*10^6 elements. It passed in 1.7s (thank god), but that's still very slow. How do people get times like 0.2s?

Don't be trapped by the title of this problem...it would just make it more complicated

Edited by author 06.02.2016 13:30

Edited by author 06.02.2016 13:38

int num,n,i,sum=0;
     // read n;

//case 1
       if(n>0){
            for(i=1;i<=n;i++){
            sum=sum+i;
        }
         printf("%d",sum);
        }
//case 2
        if(n<0){
             for(i=(-1);i>=n;i--){

            sum=sum+i;
        }
            sum=sum+1;
            printf("%d",sum);
        }
//case 3
        else if(n==0){
            printf("1");
        }

        }

Using the same code, and only switching the order of the 2 input strings ...
    scanf("%s", strr);
    scanf("%s", strl);
leads to WA5

switching the order (shouldn`t make any difference)
    scanf("%s", strl);
    scanf("%s", strr);
and I get WA6

I tried everything suggested in the discussion.
Each test works, but when I submit, I get WA2.

Can you please add more tests?

Looks like stations can overlap (have same coordinates) while not being connected with underground!