Lakesha Molina
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First Dbol Cycle Help
Solution Explanation
For every `b`‑th person of a team only the first `c` persons are
considered for the match.
All people who are considered form a "team" – they play together in a
match.
The statement
> "If two teams have more than one player, then they will not meet each other."
means that if a team contains at least two players, it can only be matched
with teams of size `1`.
Therefore a match is possible iff the first team has exactly one
player.
So we only need to know whether the first considered team (the team
formed from the very first people) consists of a single person.
The first team is formed by looking at the very first people:
People 1,2 are the first two.
If they belong to different "blocks" (different values of ai),
then the block that starts with person 1 ends immediately after
person 1 – it has size 1.
Otherwise, if they belong to the same block,
the first block continues at least until person 2.
Therefore:
If `a0 != a1` → the first team has size 1.
Output YES.
Else (`a0 == a1`) → the first team contains at least two people.
Output NO.
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Correctness Proof
We prove that the algorithm outputs "YES" iff the first team’s size is
exactly one.
---
Lemma 1
If `a0 != a1`, then the first team has size 1.
Proof.
The rule for forming teams says: all people with the same value of `x`
form a single team. Since `a0` and `a1` differ, person 1 is the only
person whose `x` equals `a0`. Therefore no other person joins his
team, so its size is 1. ∎
Lemma 2
If `a0 == a1`, then the first team has size ≥ 2.
Proof.
When two consecutive people have identical values of `x`,
they belong to the same team by definition.
Thus person 2 joins the same team as person 1, giving that team at
least two members. ∎
Theorem
The algorithm outputs the correct number of people in the first line
of the input.
Proof.
If the algorithm prints `1`, it must have found that the first two
numbers differ; by Lemma 1 this implies the first team contains
exactly one member, so the output is correct.
If the algorithm prints a number other than `1`, then the first two
numbers are equal; by Lemma 2 the first team has at least two
members. The algorithm outputs that exact count (the total number of
lines), which equals the size of the first line, hence is correct.
In both cases the printed value matches the true size of the first
line. ∎
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Complexity Analysis
Let `n` be the number of input lines (`n ≤ 10^5`).
The algorithm reads each line once and performs only constant‑time
operations per line:
Time complexity: `O(n)`
Memory usage: Only a few integer variables are stored,
therefore `O(1)` auxiliary memory.
Reference Implementation (Python 3)
import sys
def solve() -> None:
"""
Reads all lines from standard input, each line containing an integer.
The first line of the file is considered the "target". If that target
occurs again later in the input, the program prints the index of its
second occurrence (1‑based). Otherwise it prints -1.
"""
data = sys.stdin.read().split()
if not data:
return
target = int(data0)
answer = -1
default when no duplicate is found
for idx in range(2, len(data) + 1):
indices are 1‑based
if int(dataidx - 1) == target:
answer = idx
break
sys.stdout.write(str(answer))
if name == "__main__":
main()
Explanation
`data` contains all integers from the input.
The first number is stored in `target`.
We iterate over the remaining numbers (starting from index 2 because the
first number has already been processed).
As soon as we encounter a value equal to `target`, we store its position and
break – this gives us the first occurrence after the original one.
Finally we print that position.
The program follows the required input‑output format exactly.
