PROBLEM LINK:

Practice
Contest

Author:Nikhil Garg
Tester:Gerald Agapov
Editorialist:Jingbo Shang

DIFFICULTY:

Easy-Medium

PREREQUISITES:

Programming Language, Simple Math.

PROBLEM:

Perform the "Ancient Algorithm" described in the problem. And output the results in all steps. Remember that all numbers and operations are modulo by C.

EXPLANATION:

As same as the title of the problem "Magic Trick", there are some math and programming tricks in this problem.

Denote the array after i-th loop as L_i[].

The first trick is a math trick -- we can maintain a slope K_i and a intercept D_i, such that all current numbers in L_i[] equals to the original numbers K_i * L[] + D_i. For each operation, we can update the K and D as the following rules:

if operation == 'R' {
     K[i + 1] = K[i]
     D[i + 1] = D[i]
} else if operation == 'A' {
     K[i + 1] = K[i]
     D[i + 1] = D[i] + A
} else if operation == 'M' {
     K[i + 1] = K[i] * B
     D[i + 1] = D[i] * B
}

The second trick is a programming trick -- at any time, L_i[] is an interval of L[] (may reversed). Therefore, we can record the begin, end, and direction each time such that the L_i[i] equals to the L[begin]. That is,

begin = 1
end = N
direction = 1
for i = 1 to N do {
    if operation == 'R' {
        swap(begin, end)
        direction = -direction
    }
    UPDATE K and D
    print L[begin] * K + D
    begin = begin + direction
}

Beside these magic tricks, there is also a common trick of long long exceeding. That is, because C is as large as 10^18, the multiple operation may exceed the type of long long. We can use a fast multiple, similar to fast power, to solve this exceeding problem without big integer in O(logC) time.

long long multiple(long long a, long long b, long long c) // a * b % c
{
    if (b == 0) {
        return 0
    }
    long long ret = multiple(a, b >> 1, c)
    ret = (ret + ret) % c
    if (b & 1) {
        ret = (ret + a) % c
    }
    return ret
}

In summary, we can solve this problem in O(N log C) for each test case.

AUTHOR'S AND TESTER'S SOLUTIONS:

Solutions will be uploaded soon. Author's solution can be found here.
Tester's solution can be found here.