Suppose we have a number l and a monotonic increasing sequence f(m), where f(m) = am + bm [log2(m)] + cm^3 and (a = 1, 2, 3, …), (b = 1, 2, 3, …), (c = 0, 1, 2, 3, …)
Here [log2(m)] is the log to the base 2 and round the value down. so,
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. and so, on
we have to find the value m such that f(m) = l, if l is not present in the sequence then we have to print 0. We have to keep in mind that values are in such a way that they can be represented in 64 bits and the three integers a, b and c less than or equal to 100.
So, if the input is like a = 2, b = 1, c = 1, l = 12168587437017, then the output will be 23001 as f(23001) = 12168587437017
To solve this, we will follow these steps −
SMALLER_VAL := 1000000
LARGER_VAL := 1000000000000000
Define a function solve() . This will take a, b, c, n
ans := a * n
lg_val := the floor of log base 2 of n
ans := ans + b * n * lg_val
ans := ans + c * n^3
From the main method, do the following −
begin := 1
end := SMALLER_VAL
if c is same as 0, then
end := LARGER_VAL
ans := 0
while begin <= end, do
mid :=(begin + end) / 2 (take integer part only)
val := solve(a, b, c, mid)
if val is same as k, then
ans := mid
come out from the loop
otherwise when val > k, then
end := mid - 1
begin := mid + 1
Let us see the following implementation to get better understanding −
from math import log2, floor SMALLER_VAL = 1000000 LARGER_VAL = 1000000000000000 def solve(a, b, c, n) : ans = a * n lg_val = floor(log2(n)) ans += b * n * lg_val ans += c * n**3 return ans def get_pos(a, b, c, k) : begin = 1 end = SMALLER_VAL if (c == 0) : end = LARGER_VAL ans = 0 while (begin <= end) : mid = (begin + end) // 2 val = solve(a, b, c, mid) if (val == k) : ans = mid break elif (val > k) : end = mid - 1 else : begin = mid + 1 return ans a = 2 b = 1 c = 1 k = 12168587437017 print(get_pos(a, b, c, k))