Find element position in given monotonic Sequence in C++

Concept

With respect of a given integer l and a monotonic increasing sequence −

f(m) = am + bm [log2(m)] + cm^3 where (a = 1, 2, 3, …), (b = 1, 2, 3, …), (c = 0, 1, 2, 3, …) Remember,here, [log2(m)] indicates, taking the log to the base 2 and round the value down.As a result of this,

if m = 1, the value is 0.

if m = 2-3, the value is 1.

if m = 4-7, the value is 2.

if m = 8-15, the value is 3.

Our task is to determine the value m such that f(m) = l, if l doesn’t belong to the sequence then we have to print 0.

It should be noted that values are in such a way that they can be represented in 64 bits and the three integers a, b and c do not exceed 100.

Input

a = 2, b = 1, c = 1, l = 12168587437017

Output

23001
f(23001) = 12168587437017

Input

a = 7, b = 3, c = 0, l = 119753085330

Output

1234567890

Methods

Using Naive Approach − With respect of given values of a, b, c, find values of f(m) for every value of m and compare it.

Using Efficient Approach − Implement Binary Search, select m = (min + max) / 2 where min and max are indicated as the minimum and maximum values possible for m then,

If f(m) < l then increment m.
If f(m) > l then decrement m.
If f(m) = l then m is the required answer.
Now repeat the above steps until and unless the required value is found or it is not possible in the sequence.

Example

Live Demo

// C++ implementation of the approach
#include <iostream>
#include <math.h>
#define SMALL_N 1000000
#define LARGE_N 1000000000000000
using namespace std;
// Shows function to return the value of f(m) for given values of a,
b, c, m
long long func(long long a1, long long b1, long long c1, long long
m){
      long long res1 = a1 * m;
      long long logVlaue1 = floor(log2(m));
      res1 += b1 * m * logVlaue1;
      res1 += c1 * (m * m * m);
      return res1;
   }
   long long getPositionInSeries1(long long a1, long long b1,
   long long c1, long long l){
      long long start1 = 1, end1 = SMALL_N;
      // Now if c is 0, then value of m can be in order of 10^15.
      // Now if c1!=0, then m^3 value has to be in order of 10^18
      // so maximum value of m can be 10^6.
      if (c1 == 0) {
         end1 = LARGE_N;
      }
      long long ans1 = 0;
      // Now for efficient searching, implement binary search.
      while (start1 <= end1) {
         long long mid1 = (start1 + end1) / 2;
         long long val1 = func(a1, b1, c1, mid1);
         if (val1 == l) {
            ans1 = mid1;
         break;
      }
      else if (val1 > l) {
         end1 = mid1 - 1;
      }
      else {
         start1 = mid1 + 1;
      }
   }
   return ans1;
}
// Driver code
int main(){
   long long a1 = 2, b1 = 1, c1 = 1;
   long long l = 12168587437017;
   cout << getPositionInSeries1(a1, b1, c1, l)<<endl;
   long long a2 = 7, b2 = 3, c2 = 0;
   long long l1 = 119753085330;
   cout << getPositionInSeries1(a2, b2, c2, l1)<<endl;
   long long a3 = 6, b3 = 2, c3 = 1;
   long long l2 = 11975309533;
   cout << getPositionInSeries1(a3, b3, c3, l2)<<endl;
   return 0;
}