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Progressive hints first, then the full explanation and implementation when you're ready to cash out.
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AI-generated and still unreviewed. Double-check the details before internalizing them.
Hints
Open only as much as you need to keep the solve alive.
Don’t try to make the whole sequence fancy. Focus on controlling the values If you can force these directly, the problem basically folds.
A clean trick is to make each adjacent pair share exactly one “private” factor. Then the gcds are automatically different if those private factors are different.
Suppose you choose distinct pairwise coprime numbers . Can you build so that and both contain , but no other shared ?
Try Adjacent terms share exactly if the are pairwise coprime.
Use the first primes. Output Then so all gcds are distinct. Boom, done.
We want the adjacent gcds
to be all distinct. The easiest way to control a gcd is to build the common factor ourselves.
Pick distinct pairwise coprime numbers:
Primes are the obvious choice.
Now define the sequence as:
So the sequence looks like:
For the first pair:
For an internal pair :
Since all are pairwise coprime, the only shared factor is , so:
For the last pair:
Therefore the adjacent gcds are exactly
which are distinct. That is literally the entire problem.
We only need up to primes because . The -th prime is about , so any product of two consecutive primes is around , far below the limit:
So a small sieve is more than enough.
Precompute primes with a sieve once. For all test cases combined, outputting dominates:
where is the sieve limit. With , this is tiny. Memory is:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
int main() {
setIO();
const int LIM = 200000;
vector<bool> is_prime(LIM + 1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= LIM; i++) {
if (is_prime[i]) {
for (int j = i * i; j <= LIM; j += i) {
is_prime[j] = false;
}
}
}
vector<ll> primes;
for (int i = 2; i <= LIM; i++) {
if (is_prime[i]) primes.push_back(i);
}
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
ll x;
if (i == 0) {
x = primes[0];
} else if (i == n - 1) {
x = primes[n - 2];
} else {
x = primes[i - 1] * primes[i];
}
cout << x << (i + 1 == n ? '\n' : ' ');
}
}
return 0;
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
int main() {
setIO();
const int LIM = 200000;
vector<bool> is_prime(LIM + 1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= LIM; i++) {
if (is_prime[i]) {
for (int j = i * i; j <= LIM; j += i) {
is_prime[j] = false;
}
}
}
vector<ll> primes;
for (int i = 2; i <= LIM; i++) {
if (is_prime[i]) primes.push_back(i);
}
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
ll x;
if (i == 0) {
x = primes[0];
} else if (i == n - 1) {
x = primes[n - 2];
} else {
x = primes[i - 1] * primes[i];
}
cout << x << (i + 1 == n ? '\n' : ' ');
}
}
return 0;
}