CF 102280D - \u0422\u0430\u0440\u0438\u0444\u0438\u043a\u0430\u0446\u0438\u044f

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Solve time: 6m 7s
Verified: yes

Solution

Problem Understanding

A route has a value k. If a passenger wants to travel through n stops, the fare is the n-th k-gonal number. For k = 4, these are square numbers, for k = 3 they are triangular numbers, and larger values of k describe regular polygonal figures with more sides.

The input contains two integers, n and k. Here n is the number of stops the passenger wants to travel, and k determines which polygonal-number sequence the route uses. We need to output the n-th k-gonal number.

The key formula for the n-th polygonal number is

[ P_k(n)=\frac{(k-2)n^2-(k-4)n}{2}. ]

The constraints allow n to reach (10^8) and k to reach 100. The input contains only one pair of values, so the intended solution should not depend on n. A loop running once for every stop would already require up to (10^8) iterations, which is unnecessarily expensive under a 0.5 second limit. The formula gives a constant-time solution with only a few arithmetic operations.

There are two boundary cases that are easy to mishandle. First, n = 0 must produce zero. For example, the input 0 4 has output 0. A formula or implementation that assumes polygonal numbering starts from n = 1 and performs some special initialization can accidentally return 1.

Second, the division must be performed after the numerator has been formed. For example, 4 5 gives

\frac{48-4}{2}

]

A careless implementation that uses an incorrect triangular-number formula, or drops the linear term, would produce the wrong value. The square case gives another useful check: 4 4 produces (4^2=16).

Approaches

A direct approach is to construct the polygonal figure layer by layer. The first polygonal number is 1, and each following layer adds a number of points determined by k and the current layer. This is correct because polygonal numbers are defined exactly by adding successive geometric layers.

The problem is that such a simulation needs one iteration for each value up to n. With n = 10^8, the worst case is on the order of (10^8) iterations. In Python this is far beyond what a 0.5 second contest limit can reasonably allow, and even in a faster language it is solving a problem that has a closed-form expression.

The useful observation is that the layer sizes form an arithmetic progression. For a k-gonal sequence, the increase from one term to the next is

[ (k-2)n-(k-3). ]

Starting with (P_k(1)=1), summing these arithmetic increments gives a quadratic expression in n. After simplifying the sum, we obtain

[ P_k(n)=\frac{(k-2)n^2-(k-4)n}{2}. ]

The brute-force method works because it explicitly performs this summation, but it fails when n is large. The closed form performs the same summation algebraically, reducing the whole computation to constant time.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(1)	Too slow
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

1. Read n and k. They completely determine the requested polygonal number, so no additional data structure is needed.
2. Compute the numerator

[ (k-2)n^2-(k-4)n. ]

This is the standard closed form obtained by summing the arithmetic progression of layer sizes. 3. Divide the numerator by 2 and print the result. Polygonal numbers are integers, so the division is exact.

Why it works

The k-gonal sequence starts with (P_k(1)=1), and its consecutive differences form the arithmetic progression

[ k-2,\ 2(k-2)-(k-4),\ 3(k-2)-(k-4),\ldots ]

which simplifies to

[ (k-2)n-(k-3). ]

Summing these differences from the first term through the n-th term gives

\frac{n((k-2)n-(k-4))}{2}

\frac{(k-2)n^2-(k-4)n}{2}. ]

The algorithm evaluates exactly this mathematical definition, so the printed value is the required fare for every valid n and k.

Python Solution

import sys
input = sys.stdin.readline

n, k = map(int, input().split())

answer = ((k - 2) * n * n - (k - 4) * n) // 2
print(answer)

The first line reads the only input pair. There are no test cases to iterate over because the statement provides exactly one n and one k.

The expression follows the closed form directly. Computing n * n before multiplying by k - 2 keeps the correspondence with the formula clear. Python integers have arbitrary precision, so there is no overflow problem even at the maximum input size.

The // 2 operation is safe because the numerator is always even for integer n and k. Using integer division also guarantees that the output is printed as an integer rather than as a floating-point value.

The case n = 0 requires no special branch. Both terms in the numerator contain n, so the expression naturally evaluates to zero.

Worked Examples

For the first sample, n = 4 and k = 4. A value of 4 means square numbers, so the expected result is (4^2=16).

Step	n	k	(k - 2) * n * n	(k - 4) * n	Numerator	Answer
1	4	4	32	0	32	16

The formula reduces to (2n^2/2=n^2) when k = 4, confirming that the general formula agrees with ordinary squares.

For the second sample, n = 4 and k = 5. These are pentagonal numbers.

Step	n	k	(k - 2) * n * n	(k - 4) * n	Numerator	Answer
1	4	5	48	4	44	22

The result is 22, the fourth pentagonal number. The same calculation also shows why the linear correction term involving k - 4 cannot be omitted.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a fixed number of arithmetic operations are performed.
Space	O(1)	The algorithm stores only n, k, and the resulting integer.

The maximum n is (10^8), but the algorithm never iterates up to n. It performs a constant number of operations, so it easily fits the 0.5 second time limit. Python's arbitrary-precision integers also safely handle the largest possible result, which is about (4.9\cdot10^{17}) for n = 10^8 and k = 100.

Test Cases

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

def solve():
    input = sys.stdin.readline
    n, k = map(int, input().split())
    answer = ((k - 2) * n * n - (k - 4) * n) // 2
    print(answer)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout
    try:
        sys.stdin = io.StringIO(inp)
        sys.stdout = io.StringIO()
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided samples
assert run("4 4\n") == "16\n", "sample 1"
assert run("4 5\n") == "22\n", "sample 2"

# Minimum n
assert run("0 3\n") == "0\n", "zero stops"

# Minimum polygon size, triangular numbers
assert run("1 3\n") == "1\n", "first triangular number"

# Square numbers, catches the linear-term handling
assert run("5 4\n") == "25\n", "fifth square number"

# Maximum values
assert run("100000000 100\n") == "489999995200000000\n", "maximum constraints"

Test input	Expected output	What it validates
0 3	0	Minimum n and zero handling
1 3	1	First polygonal number and indexing boundary
5 4	25	Square-number special case
100000000 100	489999995200000000	Maximum constraints and large integer arithmetic

Edge Cases

The first edge case is n = 0. For input 0 4, the algorithm computes

\frac{0}{2}

]

No special condition is required, and the output is 0. An implementation that starts from the first geometric layer and assumes at least one layer exists could incorrectly return 1.

The second edge case is the first valid value, n = 1. For input 1 3, the calculation is

\frac{1+1}{2}

]

This confirms that the sequence is indexed in the standard way, with the first polygonal number equal to 1. An off-by-one implementation based on the number of added layers could accidentally produce the second triangular number instead.

The square case is another useful boundary check. For input 5 4, the formula becomes

\frac{50}{2}

]

The linear term disappears exactly when k = 4, leaving the familiar square formula.

Finally, consider the largest possible input, 100000000 100. The algorithm performs no loop at all. It evaluates

489999995200000000. ]

The result fits comfortably in Python's integer representation, and the running time remains constant despite n being as large as (10^8).