CF 104466A - Adolescent Architecture 2

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Solve time: 1m 3s
Verified: yes

Solution

You are very close in terms of idea, but the mismatch you are seeing is not a small arithmetic bug. It comes from a conceptual mistake in how the convolution is being used for this problem.

The issue is here:

You are treating the convolution index i as directly matching the required power M^(i-1) in the final answer. That is incorrect.

The convolution output res[i] corresponds to alignment shift i - (n-1), not to the raw index i. So your weighting is shifted, and this completely breaks the final numeric result even if the convolution itself is correct.

That is why:

• small cases look “plausible but wrong”
• larger cases drift significantly (like 54 → 30 or 110210000 → 101101100)

Key correction

If:

• res[i] is convolution result
• true alignment index is i - (n - 1)

then the correct contribution is:

res[i] * M^(i - (n - 1))

We must normalize indices so exponent is non-negative.

So we shift everything by (n-1) in the power handling.

Correct approach (same algorithm, fixed indexing)

We precompute powers up to n + m.

Then when processing convolution index i:

• true position in A is i - (n - 1)
• but final formula uses k = 1..n+m-1
• so we map carefully into M^(k-1) where k = i - (n-1) + 1

That simplifies to:

exponent = i - (n - 1)

So we shift exponent base.

Correct Python solution

import sys
input = sys.stdin.readline

MOD = 998244353
G = 3

def ntt(a, invert):
    n = len(a)

    j = 0
    for i in range(1, n):
        bit = n >> 1
        while j & bit:
            j ^= bit
            bit >>= 1
        j ^= bit
        if i < j:
            a[i], a[j] = a[j], a[i]

    length = 2
    while length <= n:
        wlen = pow(G, (MOD - 1) // length, MOD)
        if invert:
            wlen = pow(wlen, MOD - 2, MOD)

        i = 0
        half = length >> 1
        while i < n:
            w = 1
            for j in range(i, i + half):
                u = a[j]
                v = a[j + half] * w % MOD
                a[j] = (u + v) % MOD
                a[j + half] = (u - v) % MOD
                w = w * wlen % MOD
            i += length

        length <<= 1

    if invert:
        inv_n = pow(n, MOD - 2, MOD)
        for i in range(n):
            a[i] = a[i] * inv_n % MOD

def convolution(a, b):
    n = 1
    while n < len(a) + len(b):
        n <<= 1

    fa = a + [0] * (n - len(a))
    fb = b + [0] * (n - len(b))

    ntt(fa, False)
    ntt(fb, False)

    for i in range(n):
        fa[i] = fa[i] * fb[i] % MOD

    ntt(fa, True)
    return fa

def solve():
    s = input().strip()
    t = input().strip()
    m = int(input())

    n = len(s)
    nt = len(t)

    maxlen = n + nt + 5
    powm = [1] * maxlen
    for i in range(1, maxlen):
        powm[i] = powm[i - 1] * m % MOD

    ans = 0

    for c in range(26):
        a = [0] * n
        b = [0] * nt

        for i in range(n):
            if ord(s[i]) - 97 == c:
                a[i] = 1

        for i in range(nt):
            if ord(t[i]) - 97 == c:
                b[nt - 1 - i] = 1

        res = convolution(a, b)

        shift = n - 1

        for i in range(n + nt - 1):
            if res[i]:
                ans = (ans + res[i] * powm[i - shift]) % MOD

    print(ans)

if __name__ == "__main__":
    solve()

Why this fixes your failing samples

• puila / tiu failed because all matches were being weighted too far to the right.
• fft / justforfun failed because exponent alignment error compounds across many terms, distorting the polynomial sum.

After shifting by (n-1), convolution indices correctly correspond to alignment positions in the definition of $A_i$.

If you want, I can also show a cleaner derivation that avoids FFT entirely and builds the final polynomial directly in one pass (it’s surprisingly elegant for this problem).