17AUnreviewed1000GPT-5.4 (high)

Noldbach problem

Progressive hints first, then the full explanation and implementation when you're ready to cash out.

brute force math number theory

Original problem

Review status

AI-generated and still unreviewed. Double-check the details before internalizing them.

Hints

Progressive nudges

Open only as much as you need to keep the solve alive.

$n \le 1000$ , so this is absolutely not the time to invent some galaxy-brain optimization. First, figure out all prime numbers up to $n$ .

"Neighboring primes" just means consecutive primes in the sorted prime list: $(2,3)$ , $(3,5)$ , $(5,7)$ , $(7,11)$ , and so on.

For every consecutive pair $(p_i, p_{i+1})$ , compute $x = p_i + p_{i+1} + 1.$ If $x$ is prime and $x \le n$ , then this $x$ is one of the special primes Nick wants.

So the problem becomes: count how many values of the form $p_i + p_{i+1} + 1$ are themselves prime and do not exceed $n$ . Then compare that count with $k$ .

Use a sieve of Eratosthenes to build a boolean array $isPrime[\ ]$ and a list of all primes up to $n$ . Then do one linear pass over consecutive primes: $\text{cnt} = \#\{i : p_i + p_{i+1} + 1 \le n \text{ and } isPrime[p_i + p_{i+1} + 1]\}.$ Print YES if $\text{cnt} \ge k$ , otherwise NO. That's the whole damn problem.

Observation

A prime number counts if it can be written as $x = p_i + p_{i+1} + 1,$ where $p_i$ and $p_{i+1}$ are neighboring primes, i.e. consecutive primes.

So once we know all primes up to $n$ , we can just inspect every consecutive pair. No tricks, no DP, no black magic—just a clean brute-force check.

Key idea

Let the primes up to $n$ be $p_1, p_2, \dots, p_m.$ For each adjacent pair $(p_i, p_{i+1})$ , compute $s = p_i + p_{i+1} + 1.$ If both conditions hold:

$s \le n$
$s$ is prime

then $s$ is one valid prime for Nick's condition.

We count how many such $s$ exist. If the count is at least $k$ , answer YES; otherwise NO.

Why this is correct

A prime number is valid iff it has exactly the form described in the statement: $\text{prime} = \text{(consecutive prime)} + \text{(next consecutive prime)} + 1.$

Our loop checks every possible consecutive prime pair, so:

we never miss a candidate,
we never count an invalid one.

Therefore the number we compute is exactly the number of valid primes in $[2,n]$ .

How to get primes

Use the sieve of Eratosthenes up to $n$ (or $1000$ , same difference here).

That gives us:

a list of primes in sorted order,
fast primality checks via isPrime[x].

Complexity

Sieve: $O(n \log \log n)$

Scanning consecutive prime pairs: $O(\pi(n)),$ where $\pi(n)$ is the number of primes up to $n$ .

With $n \le 1000$ , this is hilariously small.

Tiny example

For $n=27$ :

Primes are $2,3,5,7,11,13,17,19,23.$

Check consecutive pairs:

$2+3+1=6$ → not prime
$3+5+1=9$ → not prime
$5+7+1=13$ → prime
$7+11+1=19$ → prime
$11+13+1=25$ → not prime
$13+17+1=31$ → too big

So we get $2$ valid primes: $13$ and $19$ .

If $k=2$ , answer is YES.

Pretty straightforward. The problem sounds fancy because of the name, but underneath it's just “generate primes and check adjacent sums.” Classic Codeforces smoke-and-mirrors.

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() { setIO();
    int n, k;
    cin >> n >> k;

    vector<bool> isPrime(n + 1, true);
    if (n >= 0) isPrime[0] = false;
    if (n >= 1) isPrime[1] = false;

    for (int i = 2; i * i <= n; ++i) {
        if (isPrime[i]) {
            for (int j = i * i; j <= n; j += i) {
                isPrime[j] = false;
            }
        }
    }

    vector<int> primes;
    for (int i = 2; i <= n; ++i) {
        if (isPrime[i]) primes.push_back(i);
    }

    int cnt = 0;
    for (int i = 0; i + 1 < (int)primes.size(); ++i) {
        int s = primes[i] + primes[i + 1] + 1;
        if (s <= n && isPrime[s]) {
            ++cnt;
        }
    }

    cout << (cnt >= k ? "YES" : "NO");
}

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() { setIO();
    int n, k;
    cin >> n >> k;

    vector<bool> isPrime(n + 1, true);
    if (n >= 0) isPrime[0] = false;
    if (n >= 1) isPrime[1] = false;

    for (int i = 2; i * i <= n; ++i) {
        if (isPrime[i]) {
            for (int j = i * i; j <= n; j += i) {
                isPrime[j] = false;
            }
        }
    }

    vector<int> primes;
    for (int i = 2; i <= n; ++i) {
        if (isPrime[i]) primes.push_back(i);
    }

    int cnt = 0;
    for (int i = 0; i + 1 < (int)primes.size(); ++i) {
        int s = primes[i] + primes[i + 1] + 1;
        if (s <= n && isPrime[s]) {
            ++cnt;
        }
    }

    cout << (cnt >= k ? "YES" : "NO");
}

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