CF 1044402 - Прогрессия

CF 1044402 - \u041f\u0440\u043e\u0433\u0440\u0435\u0441\u0441\u0438\u044f

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Solve time: 1m 16s
Verified: yes

Solution

Problem Understanding

Three numbers remain on the board, in their original order, after one element was erased from an arithmetic progression of four integers. We must restore the missing number and also report where it belongs so that the four resulting numbers form an arithmetic progression.

The input contains exactly three positive integers. Their order cannot be changed because they already appear in the same order as the original progression. The output consists of the missing integer and its position among the four numbers. Position 1 means the number is inserted before all given values, position 2 and 3 mean it is inserted between consecutive values, and position 4 means it is appended at the end.

The input values are at most 10^5, but there are only three of them. Since the input size is constant, any algorithm that performs only a fixed amount of work is effectively O(1). There is no need for sophisticated data structures or optimization.

Several situations require careful handling.

If the missing number is at one end of the progression, both existing differences are equal. For example,

2
4
6

The correct answer is 8 at position 4, although 0 at position 1 would also produce an arithmetic progression. A solution should accept either valid answer instead of assuming the missing value must always be inside the sequence.

If the missing number is between the first and second values, the larger gap appears first.

10
16
19

The differences are 6 and 3. The larger difference must be split into two equal differences, so the answer is 13 at position 2. Simply checking whether the two differences are equal would miss this case.

If the missing number is between the second and third values, the larger gap appears second.

10
13
19

The differences are 3 and 6. The answer is 16 at position 3. The algorithm must distinguish which gap is larger so that it inserts the value in the correct place.

If all three numbers are identical,

5
5
5

the progression has common difference 0. Any inserted value other than 5 would break the progression. The correct answer is 5 at either end.

Approaches

A straightforward idea is to try inserting a number into each of the four possible positions, construct the resulting sequence, and check whether it is an arithmetic progression. The only remaining question is which value to insert. Since every valid progression is completely determined by its common difference, we could derive candidate values from the existing numbers and verify each possibility. This approach is correct because there are only four insertion positions and a constant number of candidates.

The key observation is that with only one missing element, exactly one of three situations must occur. Either the missing number is before the first element, after the last element, or inside one of the two existing gaps. If it lies inside a gap, that gap must be exactly twice the common difference, while the other gap already equals the common difference. If it lies at an end, both observed gaps are already equal.

This observation lets us decide the answer directly by comparing the two consecutive differences.

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

Algorithm Walkthrough

1. Read the three numbers a, b, and c.
2. Compute the consecutive differences d1 = b - a and d2 = c - b.
3. If d1 == d2, the given numbers are already consecutive elements of an arithmetic progression. Insert one more element at either end. Output c + d1 at position 4.
4. If d1 > d2, the first gap is the doubled one. The missing number belongs between the first and second numbers. Output a + d2 at position 2.

The first gap equals 2 * d2, so inserting a + d2 splits it into two equal gaps of size d2. 5. Otherwise, d2 > d1. The second gap is the doubled one. Output b + d1 at position 3.

The second gap equals 2 * d1, so inserting b + d1 makes both resulting gaps equal to d1.

Why it works

The input is guaranteed to come from some arithmetic progression of four integers.

If the missing element is at one end, the remaining three numbers are consecutive elements of the progression, so both observed differences are already equal.

If the missing element is inside the progression, one observed gap skips over exactly one erased element. That gap is twice the common difference, while the other observed gap already equals the common difference. Comparing the two differences uniquely identifies which gap must be split, and inserting one common difference away from its left endpoint reconstructs the original progression.

Since these are the only possible configurations, the algorithm always returns a correct answer.

Python Solution

import sys
input = sys.stdin.readline

a = int(input())
b = int(input())
c = int(input())

d1 = b - a
d2 = c - b

if d1 == d2:
    print(c + d1)
    print(4)
elif d1 > d2:
    print(a + d2)
    print(2)
else:
    print(b + d1)
    print(3)

The program first computes the two existing gaps. Their relationship completely determines where the missing number belongs.

When the gaps are equal, the three numbers are already consecutive members of the progression, so extending the sequence at the end is always valid. Extending at the beginning would also work, but only one valid answer is required.

When one gap is larger, the input guarantee implies that it must be exactly twice the smaller gap. The inserted value is one smaller-gap distance from the left endpoint of the larger gap.

All arithmetic uses integers, so there are no rounding issues or overflow concerns.

Worked Examples

Example 1

Input:

10
16
19

Step	a	b	c	d1	d2	Decision
Read input	10	16	19	6	3	d1 > d2
Output	10	16	19	6	3	13, position 2

The first gap is twice as large as the second. Inserting 13 splits the gap 10 -> 16 into 10 -> 13 -> 16, giving equal differences of 3.

Example 2

Input:

2
4
6

Step	a	b	c	d1	d2	Decision
Read input	2	4	6	2	2	d1 == d2
Output	2	4	6	2	2	8, position 4

The three numbers already form consecutive terms of the progression, so appending 8 produces 2, 4, 6, 8.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a fixed number of arithmetic operations and comparisons are performed.
Space	O(1)	Only a few integer variables are stored.

The algorithm performs constant work regardless of the input values, so it easily satisfies any reasonable time and memory limits.

Test Cases

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

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

    a = int(input())
    b = int(input())
    c = int(input())

    d1 = b - a
    d2 = c - b

    if d1 == d2:
        print(c + d1)
        print(4)
    elif d1 > d2:
        print(a + d2)
        print(2)
    else:
        print(b + d1)
        print(3)

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()
    sys.stdin = backup_stdin
    sys.stdout = backup_stdout
    return out

# provided sample
assert run("10\n16\n19\n") == "13\n2\n"

# custom cases
assert run("10\n13\n19\n") == "16\n3\n", "missing third element"
assert run("2\n4\n6\n") == "8\n4\n", "missing last element"
assert run("5\n5\n5\n") == "5\n4\n", "zero common difference"
assert run("99998\n99999\n100000\n") == "100001\n4\n", "large values"

Test input	Expected output	What it validates
10 13 19	16, position 3	Missing element in the second gap
2 4 6	8, position 4	Missing element at the end
5 5 5	5, position 4	Zero common difference
99998 99999 100000	100001, position 4	Large input values

Edge Cases

When the missing value is before all existing numbers, for example

4
6
8

the algorithm computes d1 = d2 = 2. Since the differences are equal, it appends 10 at the end. This differs from the original progression 2, 4, 6, 8, but it is still a valid arithmetic progression, which satisfies the problem requirements.

When the missing value lies between the first and second numbers,

10
16
19

the differences are 6 and 3. The first difference is larger, so the algorithm inserts 10 + 3 = 13 at position 2. The resulting sequence is 10, 13, 16, 19.

When the missing value lies between the second and third numbers,

10
13
19

the differences are 3 and 6. The second difference is larger, so the algorithm inserts 13 + 3 = 16 at position 3. The reconstructed progression is 10, 13, 16, 19.

When every value is identical,

5
5
5

both differences are zero. The algorithm outputs another 5 at the end, producing 5, 5, 5, 5, which is an arithmetic progression with common difference 0.