### SUGGESTION:

Before posting your question with asking for help try this test.

The answer should be 436746489.

If it is hard to debug this test for you, here is the helpful information.

It contains the number of unsuccessful tries for each cell

or -1 if the cell was hacked by the “Grid Hacking Mechanism”.

If it still hard to debug try another smaller test that contains answer and similar debug information as above.

If you pass this test then read carefully the bold tip in the QUICK EXPLANATION section.

If you follow this tip but still have WA then post your question and we will help you.

### PROBLEM LINK:

**Author:** Khadar Basha

**Tester:** Anton Lunyov

**Editorialist:** Anton Lunyov

### DIFFICULTY:

EASY

### PREREQUISITES:

Connected component, Depth-first search, Flood fill algorithm, Sieve of Eratosthenes

### PROBLEM:

You are given **N × N** grid filled by non-negative integers. We single out 3 types of numbers on the grid: primes, even non-primes and odd non-primes. For each number we define the cost as follows: for the prime number it is the id of this prime in the 0-based list of all primes, for even non-prime number **X** it is **X / 2**, and for odd non-prime number **Y** it is **(Y + 3) / 2** (the cost of a number is the shortcut for the number of unsuccessful tries to crack the server secured by the password equals to this number; the mentioned formulas for cost in all three cases are clear from the problem statement).

Two even non-prime numbers that share a side on the grid are connected to each other. The same is true for odd non-prime numbers. But prime number is not connected to any other number. In this way all cells of the grid form a bidirectional graph that has several connected components (each cell having prime number is a separate component). The cost of the component is defined as the cost of the first cell in this component that we meet traversing the grid in row-major order. Now your task is to find the sum of costs over all components.

### QUICK EXPLANATION:

Actually, most of the job has already made while we reformulated the problem above. Now you simply need to loop over all cells of the grid in row-major order and if we meet the unvisited cell we add its cost to the answer and run DFS from this cell in the graph constructed above to mark other cells of its connected component as visited.

**Tip: the total cost could overflow 32-bit integer type. Use 64-bit type instead.**

To be able to find the cost of the prime cell fast enough we need to run Sieve of Eratosthenes in advance (even before input the very first test) in order to find all prime numbers. Then we need to create the array of size **10 ^{7}** that contains for each prime its id. Now the cost of each number can be found in

**O(1)**.

The overall complexity is **O(K * log log K + T * N * N)**.

### EXPLANATION:

As mentioned above at first we run Sieve of Eratosthenes to identify all prime numbers:

isPrime[0] = false isPrime[1] = false for i = 2 to K do isPrime* = true for i = 2 to sqrt(K) do if isPrime* then for j = i * i to K with step i do isPrime[j] = false

where **K = 10 ^{7} − 1**.

Next we fill array of costs. We maintain variable `prime_id`

which is equal to the number of primes we met so far:

prime_id = 0 for i = 0 to K do if isPrime* then cost* = prime_id prime_id = prime_id + 1 else cost* = i div 2 + (i mod 2) * 2

Note the formula for the cost for non-prime number.

Now we can input grids and traverse them. We use two-dimensional array `A[][]`

for the initial grid. Also we use another two-dimensional array `visited[][]`

to check visited cells, which could be filled by `false`

while input the grid:

for i = 1 to N do for j = 1 to N do read A*[j] visited*[j] = false

Now we can traverse the grid. If we meet visited cell we skip it. Otherwise we always add its cost to the answer and if it is non-prime then run DFS to fill its connected component:

total_cost = 0 // this should be a variable of 64-bit integer type! for i = 1 to N do for j = 1 to N do if (visited*[j]) then continue total_cost = total_cost + cost[A*[j]] if (not isPrime[A*[j]]) then DFS(i, j)

**DFS(i, j)** is the recursive routine that traverse the grid passing from the cell **(i, j)** to its neighbors in the graph constructed above:

DSF(i, j) if (not visited*[j]) then visited*[j] = true for (x, y) in {(i+1, j), (i-1, j), (i, j-1), (i, j+1)} do if (not isPrime[A[x][y]]) and (A[x][y] is of the same parity as A*[j]) then DFS(x, y)

Note that we pass to the cell **(x, y)** only if the number in it is non-prime and of the same parity as in the cell **(i, j)**. Missing of any of these checks will definitely lead to WA. Also see tester’s solution as a reference to one of the convenient ways how to implement loop:

for (x, y) in {(i+1, j), (i-1, j), (i, j-1), (i, j+1)} do

### ALTERNATIVE SOLUTION:

For some languages (like Java or Python) the recursive implementation of DFS could lead to runtime error due to stack overflow. In this case you need to implement either non-recursive DFS or BFS (breadth-first search) to mark cells of each connected component. Another alternative is to use disjoint-set data structure to do the same job. But this way requires some additional thinking.

The alternative very fast method to find all primes is to use Atkin Sieve invented in 2004 by Atkin and Bernstein. It allows to find all primes up to **K** using **O(K / log log K)** operations. Check it out first 6 related problems to train yourself at fast implementation of sieve.

The remaining 3 problems involves flood fill algorithm.

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.

Tester’s solution can be found here.

### RELATED PROBLEMS:

SPOJ - 6488. Printing some primes - TDPRIMES

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SPOJ - 6489. Finding the Kth Prime (Hard) - KPRIMES2

SPOJ - 3505. Prime Number Theorem - CPRIME

SPOJ - 11844. Binary Sequence of Prime Number - BSPRIME

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