CF 102697012 - Easy Exponentials

CF 102697012 - Easy Exponentials

Rating: -
Tags: -
Solve time: 2m 9s
Verified: yes

Solution

Problem Understanding

The task is to compute a small exponentiation. The input contains two integers, n and k, representing the base and exponent. The output should be the exact value of n raised to the power of k, not just a digit or a reduced form. The constraints are intentionally tiny: both values are at most 10, so the largest possible answer is only 10^10, which easily fits in common integer types. The original problem statement confirms that a direct computation is expected because the brute force approach is sufficient here.

Because the numbers are so small, there is no need for advanced exponentiation techniques, modular arithmetic, or special handling for huge values. Even a loop that multiplies the base by itself ten times performs only a constant number of operations. Any algorithm with polynomial or logarithmic complexity would also pass, but it would solve a much harder version of the problem than the one actually given.

The main edge cases come from handling the exponent correctly. If the exponent is zero, the mathematical result is one, but this input range starts from one, so that case cannot appear. The smallest valid input is a base and exponent of one. For example:

Input:
1 1

Output:
1

A careless implementation that initializes the answer incorrectly, such as starting from zero, would output zero because every multiplication would preserve that mistake.

Another possible mistake is reversing the meaning of the two numbers. The first value is the base and the second value is the exponent. For example:

Input:
2 3

Output:
8

Computing 3^2 instead would produce 9, which is wrong.

Approaches

The simplest approach is to simulate the definition of exponentiation. Starting with the answer equal to one, multiply it by the base exactly k times. This works because n^k is defined as n * n * ... * n with k copies of n. With the largest exponent being only ten, this requires at most ten multiplications.

A more general approach would be binary exponentiation, which reduces the number of multiplications for very large exponents by repeatedly squaring the base. However, the current problem does not need that optimization. Adding it would make the code longer without improving the solution under these constraints.

The brute-force method works because the exponent is bounded by a very small constant. If the exponent could be as large as 10^18, repeated multiplication would be impossible, and a logarithmic method would become necessary. Here, the structure of the input removes that need.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(k)	O(1)	Accepted
Optimal	O(k)	O(1)	Accepted

Algorithm Walkthrough

1. Read the base n and exponent k from the input. The first number controls what gets multiplied, and the second number controls how many times multiplication happens.
2. Initialize the answer as 1. This is the neutral value for multiplication, so the first multiplication produces the correct partial result.
3. Repeat k times, multiplying the current answer by n after each iteration. After the first iteration the value is n, after the second it is n^2, and after the final iteration it becomes n^k.
4. Print the final value.

Why it works: after every multiplication step, the answer stores exactly the power represented by the number of completed iterations. The loop performs exactly k multiplications, so when it finishes the stored value is precisely n^k.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, k = map(int, input().split())

    ans = 1
    for _ in range(k):
        ans *= n

    print(ans)

if __name__ == "__main__":
    solve()

The input is read directly as integers because the constraints guarantee that both values fit normally. The variable ans starts at one because multiplying by one does not change the result, which gives the correct starting point for building the power.

The loop uses range(k) so that exactly k multiplications happen. An off-by-one mistake here would be easy to make: using range(k + 1) would calculate n^(k+1), while using fewer iterations would miss a factor of n.

Python integers automatically grow when necessary, so there is no overflow concern. Even though the maximum answer here is small, using normal integer arithmetic is enough.

Worked Examples

For the input:

3 5

the execution is:

Iteration	Current answer	Operation
Start	1	Initial value
1	3	1 * 3
2	9	3 * 3
3	27	9 * 3
4	81	27 * 3
5	243	81 * 3

The final value is 243, which is 3^5. This trace shows that each iteration adds exactly one factor of the base.

For the input:

10 3

the execution is:

Iteration	Current answer	Operation
Start	1	Initial value
1	10	1 * 10
2	100	10 * 10
3	1000	100 * 10

The final value is 1000, demonstrating that the algorithm handles larger bases without any special cases.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(k)	The loop performs exactly k multiplications.
Space	O(1)	Only the current answer and input values are stored.

Since k is at most 10, the number of operations is constant and the solution easily fits within the given limits.

Test Cases

import sys
import io

def solution(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)

    n, k = map(int, sys.stdin.readline().split())
    ans = 1
    for _ in range(k):
        ans *= n

    sys.stdin = old_stdin
    return str(ans)

# provided sample
assert solution("3 5\n") == "243", "sample 1"

# minimum values
assert solution("1 1\n") == "1", "minimum input"

# base equal to maximum value
assert solution("10 10\n") == "10000000000", "maximum input"

# same values
assert solution("5 5\n") == "3125", "all equal values"

# boundary where reversing base and exponent changes result
assert solution("2 10\n") == "1024", "base exponent order"

Test input	Expected output	What it validates
3 5	243	Provided sample behavior
1 1	1	Smallest valid input
10 10	10000000000	Maximum constraints
5 5	3125	Equal base and exponent
2 10	1024	Correct interpretation of base and exponent

Edge Cases

For the smallest possible input:

1 1

the algorithm starts with ans = 1, performs one multiplication, and gets 1. This confirms that the initialization does not accidentally introduce a zero result.

For a case where the order of the two numbers matters:

2 3

the loop multiplies by 2 three times:

1 -> 2 -> 4 -> 8

The output is 8. A mistaken implementation that treated the first number as the exponent would compute 3^2 and fail this case.

For the largest allowed input:

10 10

the loop performs ten multiplications and reaches 10000000000. The algorithm does not need special handling because Python integers can represent the result exactly.