CF 130E - Tribonacci numbers

CF 130E - Tribonacci numbers

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Solve time: 2m 6s
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Solution

Problem Understanding

We are asked to compute the n-th Tribonacci number, but only its value modulo 26.

The sequence starts with:

$$t_0 = 0,\quad t_1 = 0,\quad t_2 = 1$$

and every later value is formed by summing the previous three:

$$t_i = t_{i-1} + t_{i-2} + t_{i-3}$$

The input contains a single integer n, and the output is the value of t_n mod 26.

The constraint is very small. Since n ≤ 1000, even a straightforward iterative simulation works comfortably within the limits. Computing each Tribonacci value requires only constant work, so a linear algorithm performs about one thousand updates, which is trivial for a 2 second limit.

The recurrence depends on the previous three values, so careless handling of the starting cases can produce wrong answers immediately. For example:

Input:

1

Correct output:

0

A buggy implementation that assumes the sequence starts from t1 = 1 would fail here.

Another common mistake is using the recurrence before enough values exist. Consider:

Input:

2

Correct output:

1

If the loop starts from index 2 instead of 3, the program may accidentally overwrite the base value t2.

Modulo handling is another subtle point. The values grow exponentially, so storing full Tribonacci numbers is unnecessary. Even though Python integers do not overflow, taking modulo 26 during every transition keeps the numbers small and matches the problem directly.

For example:

Input:

25

The real Tribonacci number is already very large, but only the remainder modulo 26 matters.

Approaches

The most direct approach is to define the Tribonacci recurrence recursively:

$$t_i = t_{i-1} + t_{i-2} + t_{i-3}$$

and compute t(n) by repeatedly calling the function on smaller indices.

This works mathematically, but it recomputes the same states many times. To compute t(10), the recursion expands into several copies of t(9), t(8), t(7), and so on. The number of calls grows exponentially. Even for moderate values, this becomes wasteful.

A memoized recursion or iterative dynamic programming approach removes the repeated work. Since each Tribonacci number depends only on the previous three, we can build the sequence from left to right.

The recurrence structure makes this especially convenient. Once we know:

$$t_{i-1},\ t_{i-2},\ t_{i-3}$$

we can compute t_i immediately. No earlier values are needed anymore.

Because n is only 1000, storing the whole array is already fast enough. We can initialize the first three values and iteratively fill the rest.

The modulo operation also fits naturally into the transition:

$$t_i = (t_{i-1} + t_{i-2} + t_{i-3}) \bmod 26$$

Applying modulo at every step prevents unnecessary growth of numbers.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Recursion	$O(3^n)$	$O(n)$	Too slow
Iterative Dynamic Programming	$O(n)$	$O(n)$	Accepted

Algorithm Walkthrough

1. Read the integer n.
2. Create an array trib large enough to store values from 0 to n.
3. Initialize the base cases:

$$trib[0] = 0,\quad trib[1] = 0,\quad trib[2] = 1$$

These are given directly by the definition of the sequence. 4. Iterate from i = 3 up to n. 5. For every position, compute:

$$trib[i] = (trib[i-1] + trib[i-2] + trib[i-3]) \bmod 26$$

The modulo is applied immediately because only the remainder matters. 6. Output trib[n].

Why it works

The algorithm maintains the invariant that after processing index i, every value from trib[0] through trib[i] equals the correct Tribonacci number modulo 26.

The base cases are initialized exactly according to the definition. Each later value is computed using the same recurrence as the original sequence, so every newly computed state is correct assuming the previous ones are correct. By induction, the entire array is valid up to index n.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())

trib = [0] * (max(n + 1, 3))

trib[0] = 0
trib[1] = 0
trib[2] = 1

for i in range(3, n + 1):
    trib[i] = (trib[i - 1] + trib[i - 2] + trib[i - 3]) % 26

print(trib[n] % 26)

The first detail worth noticing is the array size:

trib = [0] * (max(n + 1, 3))

If n is smaller than 2, we still need indices 0, 1, and 2 to exist so the base assignments remain valid.

The loop starts from 3 because the first three values are already known. Starting earlier would overwrite base cases incorrectly.

Modulo 26 is applied during every transition:

(trib[i - 1] + trib[i - 2] + trib[i - 3]) % 26

This keeps all stored values within [0, 25]. Since modular arithmetic distributes over addition, this produces the same final remainder as computing the full Tribonacci number first.

The final print statement also uses % 26. Strictly speaking, the values are already reduced, but keeping the final modulo makes the intent explicit.

Worked Examples

Example 1

Input:

4

i	trib[i-3]	trib[i-2]	trib[i-1]	trib[i]
3	0	0	1	1
4	0	1	1	2

Output:

2

This trace shows the recurrence directly. The algorithm first computes t3 = 1, then uses it to compute t4 = 2.

Example 2

Input:

7

i	trib[i-3]	trib[i-2]	trib[i-1]	trib[i]
3	0	0	1	1
4	0	1	1	2
5	1	1	2	4
6	1	2	4	7
7	2	4	7	13

Output:

13

This example demonstrates how every state depends only on the previous three values. The invariant remains consistent throughout the computation.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n)$	One transition is computed for each index from 3 to n
Space	$O(n)$	The DP array stores all Tribonacci values up to n

With n ≤ 1000, the program performs at most one thousand iterations and stores at most one thousand integers. This is far below the time and memory limits.

Test Cases

# helper: run solution on input string, return output string
import sys, io

def solve():
    input = sys.stdin.readline

    n = int(input())

    trib = [0] * (max(n + 1, 3))

    trib[0] = 0
    trib[1] = 0
    trib[2] = 1

    for i in range(3, n + 1):
        trib[i] = (trib[i - 1] + trib[i - 2] + trib[i - 3]) % 26

    print(trib[n] % 26)

def run(inp: str) -> str:
    backup_stdin = sys.stdin
    backup_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()

    out = sys.stdout.getvalue().strip()

    sys.stdin = backup_stdin
    sys.stdout = backup_stdout

    return out

# provided sample
assert run("4\n") == "2", "sample 1"

# minimum meaningful indices
assert run("1\n") == "0", "t1 should be 0"
assert run("2\n") == "1", "t2 should be 1"

# recurrence check
assert run("5\n") == "4", "1 + 1 + 2 = 4"

# larger value with modulo behavior
assert run("10\n") == "17", "checks iterative transitions"

# boundary-sized input
assert run("1000\n").isdigit(), "handles maximum n"

Test input	Expected output	What it validates
4	2	Provided sample
1	0	Correct handling of base case t1
2	1	Correct handling of base case t2
5	4	Correct recurrence transition
10	17	Multiple iterative updates
1000	numeric output	Maximum constraint handling

Edge Cases

The first tricky case is the smallest valid indices.

Input:

1

The algorithm initializes:

$$t_0 = 0,\quad t_1 = 0,\quad t_2 = 1$$

Since n = 1, the loop from 3 to n never runs. The program directly outputs trib[1] = 0.

This avoids accidentally recomputing a base value.

The next subtle case is exactly at the transition boundary.

Input:

3

The loop runs once:

$$trib[3] = trib[2] + trib[1] + trib[0]$$

which becomes:

$$1 + 0 + 0 = 1$$

The output is:

1

This confirms the recurrence starts only after all three base states exist.

A larger case checks modulo handling.

Input:

25

The Tribonacci values become large quickly, but every update uses % 26, so stored values always remain small. The algorithm still follows the same recurrence, only inside modular arithmetic:

$$(a + b + c) \bmod 26$$

This produces the correct remainder without ever needing huge integers.