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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
// 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;
}Output
23001 1234567890 0
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