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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Strange Counter
  5. Discussions

Strange Counter

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983 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/Plsx1i8dqiI

    long strangeCounter(long t) {
    long last = 3, step = 3;
    while(t > last){
        step *= 2;
        last += step;
    }
    return last - t + 1;
}

  • Theo_Upton
     + 0 comments

    C++ solution:

    long strangeCounter(long t) {
    return 6*pow(2,floor(log2((1+(t-1)/3))))-2-t;
}

  • zaidnsour12
     + 0 comments

    Java solution:

    public static long cycleLength(long t){
    long sum = 0;
    long len = 3;
    while(true){
        sum = sum +len;
        if(sum >= t) return len;
        len = len * 2;
        }      
    }
    
  public static long strangeCounter(long t) {
    return 2 * cycleLength(t) - t - 2;
    }

  • alex_k5
     + 0 comments

    def strangeCounter(t): sm = 0 for i in range(10**12): sm += 2**i if 3*sm >= t: return 3*sm-t+1

  • tangjiaqing
     + 0 comments

    This is the full steps of calculation:

    public static long Run(long time)
{
// a[k] = a[k-1] * q
// a[n] = a[m] * q^(n-m)
// S = a[1] * (1 - q^n) / (1 - q)

var S = new Dictionary<long, long>();
var a = new Dictionary<long, long>();

a[1] = 3;

var q = 2;

var n = (long)Math.Ceiling(Math.Log(1 - time * (1 - q) / (double)a[1]) / Math.Log(q));

S[n - 1] = (long)(a[1] * (1 - Math.Pow(2, n - 1)) / (1 - q));
a[n] = a[1] * (long)Math.Pow(q, n - 1);

var remain = time - S[n - 1];
var val = a[n] - remain + 1;

return val;
}