Noldbach 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.