#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define SIZE 128*1024
int main()
{
    int i,n=0;
    long long int num[SIZE],N;

printf("enter your aray value:\n");
    while(scanf("%lld",&N)!=EOF)
          {
              num[n]=N;
            n++;
        }

    for(i=n-1;i>=0;i--){

      printf("%0.4lf\n",(double)sqrt(num[i]));
    }
    return 0;
}

4th test is:

3
0 0
9999 10000
9998 9999

from re import *
from math import sqrt

result = findall('[0-9]+', input())
result.reverse()

for i in result:
    print('{:.4f}'.format(sqrt(int(i))))

Edited by author 14.03.2017 01:20

Who had a Wrong Answer in 5 th test ? What mistake may be there ?

Could someone provide tests for me? Thank you

Try this test:
4 3
####
#..#
####
Answer is 1 :)

#include <bits/stdc++.h>

using namespace std;

int main()
{int x;
     vector <string> myvector (2001);
  fill (myvector.begin(),myvector.begin()+5,"few");
  fill (myvector.begin()+6,myvector.begin()+10,"several");
  fill (myvector.begin()+11,myvector.begin()+20,"pack");
  fill (myvector.begin()+21,myvector.begin()+50,"lots");
  fill (myvector.begin()+51,myvector.begin()+100,"horde");
  fill (myvector.begin()+101,myvector.begin()+250,"throng");
  fill (myvector.begin()+251,myvector.begin()+500,"swarm");
 fill (myvector.begin()+501,myvector.begin()+1000,"zounds");
 fill (myvector.begin()+1001,myvector.end(),"legion");
 cin>>x;
 cout<<myvector[x];

    return 0;
}
correct solution:
#include <bits/stdc++.h>

using namespace std;

int main()
{int x;
     vector <string> myvector (2001);
  fill (myvector.begin(),myvector.begin()+5,"few");
  fill (myvector.begin()+5,myvector.begin()+10,"several");
  fill (myvector.begin()+10,myvector.begin()+20,"pack");
  fill (myvector.begin()+20,myvector.begin()+50,"lots");
  fill (myvector.begin()+50,myvector.begin()+100,"horde");
  fill (myvector.begin()+100,myvector.begin()+250,"throng");
  fill (myvector.begin()+250,myvector.begin()+500,"swarm");
 fill (myvector.begin()+500,myvector.begin()+1000,"zounds");
 fill (myvector.begin()+1000,myvector.end(),"legion");
 cin>>x;
 cout<<myvector[x]<<endl;;

    return 0;
}

Edited by author 30.09.2017 07:20

I want to know how to approach this question in what direction to think so that i can solve it on my own.
Thank you.

Маяки могут находится в 1 точке.
P.S.
 Если считать, что маяки не могут быть в одной точке, то тогда нужно решать паросочетанием (как я в начале и решал), что гораздо сложнее по реализации

I tried different tasks and always got the right answer.

any test?

What's wrong in my code?
<
#include <stdio.h>
#include <string.h>

int main() {
        int i,h,w,n,tmp;
    int curLen = 0, curNumStr = 0;
    char s[105];
    scanf("%d%d%d\n",&h,&w,&n);
    for(i = 0; i < n; i ++) {
        gets(s); //scanf("%s",s);
        tmp = strlen(s);
        if(curLen)
            curLen += 1 + tmp;
        else
            curLen += tmp;
        if(curLen / w){
            curNumStr += curLen / w;
            if(curLen % w)
                curLen = tmp;
            else
                curLen = 0;
        }
    }
    if(curLen)
        curNumStr ++;

    tmp = curNumStr / h;
    if(curNumStr % h)
        tmp ++;
    printf("%d",tmp);
    return 0;
}
>

I got WA at test 18. Any tests for me?

Hello,
I didn't understand the problem statement correctly. Do cities having power stations need to be connected to each other? The solution can be also having more than 1 connected components( Graphs ), right?

Edited by author 21.09.2017 09:28

Edited by author 21.09.2017 09:32

Hello,
I am not that strong in DP but I have been trying hard to understand the problem. I tried coming up with a solution however got WA. I cannot completely understand the hints given before, PLEASE HELP !!!

My solution is very ugly
Maybe yours is beautiful
How nice is your solution?

const maxH=500;
var N,K,q,w,e:longint;
    H:array[1..maxH] of longint;
    z:array[1..2,1..maxH,0..2] of longint;
    x:array[0..1] of longint;

begin
readln(N,K);
for q:=1 to N do
  readln(h[q]);

for q:=1 to K do
  begin
  z[2,q,0]:=0;
  z[2,q,1]:=0;
  z[2,q,2]:=0;
  end;
z[2,1,h[1]]:=1;

for q:=2 to N do
  begin
  z[1]:=z[2];
  for e:=1 to K do
    begin
    z[2,e,0]:=0;
    z[2,e,1]:=0;
    z[2,e,2]:=0;
    end;
  z[2,1]:=z[1,1];
  z[2,1,h[q]]:=z[2,1,h[q]]+1;

  e:=K;
  if q<K then e:=q;
  for w:=2 to e do
    begin
    x[0]:=z[1,w,0];
    x[1]:=z[1,w,1];
    x[h[q]]:=x[h[q]]+1;

    z[2,w,h[q]]:=1;
    z[2,w,2]:=z[1,w-1,0]*z[1,w-1,1]+z[1,w-1,2];

    if z[2,w,2]>z[1,w,2]+x[0]*x[1] then
      begin
      z[2,w,0]:=x[0];
      z[2,w,1]:=x[1];
      z[2,w,2]:=z[1,w,2];
      end;
    end;
  end;
writeln(z[2,K,0]*z[2,K,1]+z[2,K,2]);
end.

ответьте на вопрос:
допустим есть цилиндр с  размеромами: 1 на 1000
и отверсие в стене размером 2 на 4,
можно ли нарезать мелкими лоскутками цилиндр на размеры 1 на 2, таких отрезков понадобиться 4 штуки, подобный ход удовлетворяеет поставленной задаче ?