Try not to use double, probably there is something like ceil(log(n)) = ceil(1.000000000000000000000001) = 2.0

Also in test 6 N is greater than 1.000.000.

I used Int and Float data types and got AC

Hint:
we start from kk = 1, then kk += kk (2), kk += kk (4) but we cannot advance when kk > k.
get to the maximum kk with this and the remaining n - kk can be easily calculated with
(n - kk) / k + (((n - kk) % k) > 0)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main() {

   int n, k;

   scanf("%d %d", &n, &k);

   if (n < 1 || n > pow(10, 9) || k < 1 || k > pow(10, 9)) {
      return 0;
   }

   unsigned long long result = 0;
   unsigned long long buffer = 1;
   n--;

   for (int i = 0; n >= 0; i++) {

      n -= buffer;
      result++;
      buffer *= 2;

      if (n <= 0) {
         printf("%llu\n", result);
         return 0;
      }

      if (buffer >= k) {
         buffer = k;
         double count = (double)n / buffer;
         if (count != (long long)count) {
            count++;
         }
         result += count;
         printf("%llu\n", result);
         return 0;
      }
   }
   return 0;
}

Edited by author 21.01.2020 01:57

Edited by author 23.06.2021 07:43

Edited by author 23.06.2021 07:43

Edited by author 23.06.2021 07:43

The program is already installed in one computer.
Now, we need to install it in (N-1) computers.

if k > n my program worked with n < 0
you should to consider the case after cycle: if n < 0 then n = 0

Edited by author 03.04.2020 21:51

#include<stdio.h>
#include<math.h>

int main()
{
    int n,k,i=1,s=1,t=0,temp=0;
    scanf("%d %d",&n,&k);

    while(i<k)
    {
        s=s+i;
        i=2*i;
        t++;

        if(s>=n)
            break;
    }
    while(s<n)
    {
        s=s+k;
        t++;
    }

    printf("%d",t);
    return 0;
}

import sys
n,k=map(int,input().split())
ans=0
nn=1
on=1
if n==1:
    print(0)
    sys.exit()
for i in range(100000000000000000000):
    if on<=k:
        nn+=on
        on+=on
        ans+=1
    elif k<n:
        nn+=k
        on+=k
        ans+=1
    if nn>=n:
        print(ans)
        sys.exit()
i can not understand why optimization my cod
if i got ac with you help i give you 1000 rubles

Edited by author 17.10.2018 22:35

n = 1

var n,k,ch,im:int64;
begin
read(n,k);
n:=n-1;
ch:=0;
im:=1;
while n>0 do begin
  if im<=k then begin
    n:=n-im;
    im:=im*2;
    ch:=ch+1;
    end;
  if im>k then begin
    n:=n-k;
    im:=im+k;
    ch:=ch+1;
    end;
  end;
writeln(ch);
end.

#include <iostream>

using namespace std;

int main()
{
    long int n, k;
    cin >> n >> k;
    long int a = 1;
    long int h = 0;
    if(n == 1){
        cout << 0;
    }
    else if(n == 2){
        cout << 1;
    }
    else{
    while(n != 0){
            if(a > k && a < n){
                a = k;
        n -= a;
        h++;
        }
        else if(a > n){
            n = 0;
            h++;
        }

        else{
            n -= a;
        a *= 2;
        if(a > k){
            a = k;
        }
        h++;
        }
    }
    cout << h;
}
}

I tried to pick up to my occasion all data types ,but there was some problem ;Then i  tried using data type "auto" and it's worked!

n, k =map(int, input().split(' '))
z=0
time=0
num = 1
progress = 2
if k==1 and n==1:
    print(n)
elif k ==1:
    print(n-1)
elif k>=n:
    time+=1
    while num<n:
        num+=progress
        progress*=2
        time+=1
    print(time)
else:
    time+=1
    while z==0:
        if progress<=k and num<n:
            num+=progress
            progress*=2
            time+=1
        else:
            z+=1
    progress=k
    if num<n:
        while num<n:
            if num<n:
                time+=1
                num+=progress
    print(time)

Edited by author 16.04.2019 01:34

I can't find mistake, help!!


#include <iostream>
#include <cmath>
using namespace std;
int main() {
    __int64 n,k;
    int s = 0;
    cin>>n>>k;
    int step;
    step = log(k)/log(2);
    if(n-pow(2,step+1)>=0) {
        n -= pow(2,step+1);
        s = (step+1)+ceil(n*1./k);
    } else {
        s = ceil(log(n)/log(2));
    }
    cout<<s;
}

Formula:

ceil(log2(min(n, k))) + ceil((double)(n - (1 << (int)(ceil(log2(min(n, k))))))/ k)

Edited by author 29.07.2018 13:01

Edited by author 09.09.2018 17:50

Here is my code:
int main()
{
    int N, K;
    cin >> N >> K;
    int connect = 1;
    int temp = N;
    int largepow = 0;
// calculate the largest power of 2 less than K
    while(temp > 0)
    {
        if(pow(2, largepow) < K && pow(2, largepow + 1) < K)
            largepow++;
        temp /= 2;
    }
    N--;
    int count = 0;
// count the hours
    while(N > 0)
    {
        N -= connect;
        count++;
        if(connect < pow(2, largepow))
            connect *= 2;
        else if(connect >= pow(2, largepow) && connect < K)
            connect++;
    }
    cout << count;
    return 0;
}

Got WA4, then I tried something else:

int main ()
{
    long long n, k, exp;
    cin >> n >> k;
    for (exp = 0; (1 << exp) < n && (1 << exp) < k; exp++)
        if ((1 << exp) < n)
            exp += ((n - (1 << exp) - 1) / k) + 1;
    cout << exp;
    return 0;
}
Test case 4 again. What is wrong?

where is my mistake?

#include <cstdio>
#include <cmath>

inline int lb(int n) { return ceil(log(double(n))/log(2.0)); }

int main() {
    int n,k;
    scanf("%d%d",&n,&k);
    if (n<=k) printf("%d",lb(n));
    else {
        int ans=lb(k);
        n-=1l << ans;
        ans+=n/k;
        if (n%k) ++ans;
        printf("%d",ans);
    }
}

What is a test 6?

If n = 1 answer is 0,  but accepted solution with answer 1.
Why???