How to find the Nth number in a fractal sequence? - c++

Question

The assignment is to write a C++ program which takes the input number n and outputs the nth number in the sequence:
1 1 2 1 2 3 1 2 3 4 1 2 3 4 5 1 2 3 4 5 6 ...
This is what I've come up with so far:
#include <iostream>
using namespace std;
int main()
{
long long n,k=1,result;
cin >> n;
if(n==1){
result=1;
}else{
for(int i=1,j=1;;i=j,j=j+k){
if(n>i&&n<=j){
result=n-i;
break;
}else{
k++;
}
}
}
cout << result << endl;
}
This is also what I've written before:
#include <iostream>
using namespace std;
int main()
{
long long n,count=0,result;
cin >> n;
for(int i=1;;i++){
for(int j=1;j<=i;j++){
count=count+1;
if(count==n){
result=j;
break;
}
}
if(count>=n){
break;
}
}
cout << result << endl;
}
Both of these work properly for smaller numbers, but the problem is I have to follow the constraint:
1 <= n <= 10^12
So when bigger numbers are inputted, the programs both take too long to output the solution and exceed the time limit, which is 2 seconds. I've been working on this for 5 hours now and I don't know how to improve these programs so they are faster. I also thought about a certain formula that could help determine the nth number in such a sequence, but I can't seem to find anything about it on the internet or in my math books. Could somebody point me to the solution? I would be very grateful.

Answer 1

We can group numbers in your sequence:
(1) (1, 2) (1, 2, 3) ...
The overall amount of numbers is
1 + 2 + 3 + ...
The latter is an arithmetic progression, its sum equals to x*(x+1)/2.
We'll find the number of full groups k that go before n+1-th number in the sequence. k equals to the maximal integer such that k*(k+1)/2 <= n. To find it we'll solve the quadratic equation:
x*(x+1)/2 = n
x^2 + x - 2*n = 0
Let's assume that positive root of this equation is x'. We round it down to the nearest integer k. If x' == k (x' is a whole number) it is the answer. Otherwise, the answer is n - k*(k+1)/2.
Exemplary c++ implementation:
double d = 1 + 8.0 * n;
double x = (-1 + sqrt(d)) / 2;
long long k = floor(x);
long long m = k*(k+1) / 2;
if (m == n) {
return k;
} else {
return n - m;
}
The solution has O(1) time complexity.

Answer 2

The first job is to write out the sequence like this:
1
2 3
4 5 6
7 8 9 10
And note that we want to map this to
1
1 2
1 2 3
1 2 3 4
The row position of a number is given by rearranging the formula for an arithmetic progression, solving the resultant quadratic, discarding the negative root, and removing any fractional part of the answer. A number t appears in the row r given by the whole number part
r = R(1/2 + (1/4 + 2 * (t - 1))1/2)
Where R() is a function that rounds a number downwards to the whole number.
But you are after the column c. That is obtained subtracting the value of the first term in that row from t:
c = t - 1/2 * r * (r - 1)
Reference: https://en.wikipedia.org/wiki/Arithmetic_progression

Answer 3

A solution using loop. It will out the number at nth.
x = 0 ;
i = 1 ;
do {
x += i ;
if( x == n ) {
cout<< i ;
break ;
}
else if( x > n ) {
cout<< (n - (x-i)) ;
break ;
}
i ++ ;
}while( 1) ;