
Н.Өнөрцэцэгийн шүүхийн шийдвэрийн тухай Сүхбаатар дүүргийн Эрүүгийн хэргийн анхан шатны шүүхийн шийдвэр:
Сүхбаатар дүүргийн Эрүүгийн хэргийн анхан шатны шүү
Шийдвэрийн дугаар: 2024.07.19 №2024/ШЦТ/707
Шийдвэрийн төлөв: Хүчин төгөлдөр болоогүй.
Шүүгдэгч Н.Өнөрцэцэгийг “Нэр бүхий 4 иргэн, 1 хуулийн этгээдийн нэр төр, алдар хүнд, ажил хэргийн нэр хүндэд халдсан илт худал мэдээллийг олон нийтэд тараасан”, “Хувь хүний хуулиар хамгаалагдсан нууцыг олж мэдсэн хүн өөрийнх нь зөвшөөрөлгүйгээр цахим сүлжээ ашиглаж задруулсан”, “Татвар төлөгч хувь хүн, хуулийн этгээдийн удирдах, гүйцэтгэх албан тушаалтан татвар төлөхөөс зайлсхийх зорилгоор их хэмжээний татвар ногдох орлогоо зориуд худал тодорхойлсон, нуусан”, “Татвар төлөхөөс зайлсхийх гэмт хэргийн улмаас олсон хөрөнгө, мөнгө, орлого гэдгийг мэдсээр байж ашигласан”, “Төрийн нууцад хамаарах мэдээ, баримтыг хадгалсан” гэмт хэрэг үйлдсэн гэм буруутайд тус тус тооцож, нийт 4 жил 9 сарын хугацаагаар хорих ялаар шийтгэж, оногдуулсан ялыг нээлттэй хорих байгууллагад эдлүүлэхээр шийдвэрлэв.
Шүүгдэгч Н.Өнөрцэцэгт холбогдох хэрэг нь төрийн нууц болон хүний эмзэг мэдээлэлтэй холбоотой тул хуульд зааснаар шүүх хуралдааныг бүхэлд нь хаалттай явуулсан болно.
Нийслэлийн Прокурорын газраас яллах дүгнэлт үйлдэж ирүүлсэн шүүгдэгч Н.Өнөрцэцэгт холбогдох хэргийн шүүх хуралдааныг 2024 оны 07 дугаар сарын 18-ны өдрийн 13 цаг 30 минутад зарласан.
Уг шүүх хуралдааныг шүүгдэгчийн өмгөөлөгч Б.Баатарсайхан нь “Нийслэлийн Эрүүгийн хэргийн давж заалдах шатны шүүх болон Сүхбаатар дүүргийн Эрүүгийн хэргийн анхан шатны шүүхэд шүүх хуралдаан давхацсан тул хойшуулах” хүсэлтийг ирүүлсэн.
Өмгөөлөгч Б.Баатарсайханаас ирүүлсэн хүсэлтэд дурдсан Нийслэлийн Эрүүгийн хэргийн давж заалдах шатны шүүхийн хуралдаан хойшилсон бол шүүгдэгч Н.Өнөрцэцэгт холбогдох хэргийн шүүх хуралдааныг 2024 оны 07 дугаар сарын 01-ний өдөр хойшлуулж, тухайн өдрөө дараагийн шүүх хуралдааны товыг 2024 оны 07 дугаар сарын 18-ны өдөр хийхээр зарласан бөгөөд өмгөөлөгчийн Сүхбаатар дүүргийн Эрүүгийн хэргийн анхан шатны шүүхэд зарлагдсан хурал нь уг хурлаас хойш зарлагдсан хурал байсан ба өмгөөлөгч О.Батсүх нь ямар нэгэн баримтыг шүүхэд ирүүлэлгүй, шүүх хуралдаанд хүндэтгэн үзэх шалтгаангүйгээр хүрэлцэн ирээгүй тул шүүгдэгч Н.Өнөрцэцэгийн өмгөөлөгч нарын зүгээс хэрэг хянан шийдвэрлэх ажиллагаанд саад учруулсан гэж үзсэн болно.
Мөн шүүх хуралдаанд шүүгдэгчийн өмгөөлөгч Д.Оросоо нь хүрэлцэн ирсэн тул шүүгдэгч Н.Өнөрцэцэгийн өмгөөлүүлэх, хууль зүйн туслалцаа авах эрх нь хангагдсан гэж дүгнэн өмгөөлөгч Б.Баатарсайханаас гаргасан шүүх хуралдааныг хойшлуулах тухай хүсэлтийг хүлээн авахаас татгалзаж шийдвэрлэсэн.
Улмаар шүүх хуралдааны хэлэлцүүлэг эхлэхээс өмнө шүүгдэгч Н.Өнөрцэцэг нь өмгөөлөгч Д.Оросоогоос өөрийн хүсэлтээр татгалзаж байгаа талаарх хүсэлтийг шүүхэд гаргасан ба шүүгдэгч нь өмгөөлөгчөөсөө татгалзах, өөрийгөө өмгөөлөх эрх нь түүний хуулиар олгогдсон эрх тул түүний хүсэлтийг хүлээн авч, хуульд заасан заавал өмгөөлөгчтэй оролцох тохиолдол биш учир шүүх хуралдааныг цааш үргэлжлүүлсэн болно.
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Unser Evogene Somatropin Evogene HGH Alley Wachstumshormone als synthetische HGH Alley wird aber auch gerne als angebliches Anti-Aging-Mittel und als Dopingmittel eingenommen.Für den Nutzen als Anti-Aging-Mittels gibt es keine Nachweise.
Das Peptid kann ohne Angst vor Nebenwirkungen langfristig verwendet werden. In klinischen Studien wurde festgestellt, dass HGH Frag außer
der bereits erwähnten Stimulierung der Lipolyse und der Hemmung der Lipogenese
keine weiteren positiven oder negativen Auswirkungen des Wachstumshormons hat.
Zudem können Verbraucher, optimistic Studien einsehen, die belegen, dass es sich bei dem Produkt Trenorol, um einen funktionierenden Testosteron Booster,
als Nahrungsergänzung handelt, der aus nur natürlichen Inhaltsstoffen zusammengesetzt ist.
D-Bal ist nicht nur ein qualitativ hochwertiges Produkt von der Crazy Bulk GmbH, es hinterlässt unter Garantie kein Nebenwirkungen und somit keine Schäden, wenn es sich um die Gesundheit des Anwenders handelt.
Bei TestoPrime handelt es sich um einen rezeptfreien Testosteron Booster,
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Dem gegenüber steht ein Missbrauch von Steroiden, die zu vielen gesundheitlichen Problemen führen können, wie zum Probleme an der Bauchspeicheldrüse(2).
Machen Sie einen Schritt zur Optimierung Ihres Wohlbefindens und entfalten Sie
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Nebenwirkungen die extrem selten bis gar nicht auftreten sind
die mögliche Stimulierung des Wachstums von Krebstumoren.
Wenn zusätzlich Protein gegessen wird, kann der Protein Wert schnell
zu viel werden. Dies ist teilweise auf Flüssigkeitsansammlungen (Ödeme) zurückzuführen,
aber auch Taubheitsgefühle und Hautreizungen sind häufig.
Die meisten auf dem Markt erhältlichen Präparate sind
im Allgemeinen für den Menschen sicher, aber nicht frei von Nebenwirkungen.
Wachstumshormon HGH (Somatotropin oder Bodybuilding HGH) ist ein Protein, das aus der
191. Die Produktion und Sekretion dieses HGH erfolgt im Hypophysenvorderlappen. Im Körper wird das Wachstumshormon in der größten Menge produziert als alle existierenden Hypophysenhormone.
Es ist kein Geheimnis, dass in Sportarten mit hohen Anforderungen, einschließlich professionellem Bodybuilding, großartige Leistungen im weitesten Sinne erforderlich sind.
Die Entscheidung, HGH in Ihre Bodybuilding-Strategie aufzunehmen, sollte nach einem umfassenden Verständnis sowohl der potenziellen Vorteile als auch der damit verbundenen Risiken getroffen werden. Indem Sie einem ausgewogenen Ansatz den Vorzug geben und professionelle Beratung einholen, können Sie fundierte
Entscheidungen treffen, die Ihre Bodybuilding-Ziele unterstützen und
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und was bei einem Sportler funktioniert, ist für einen anderen möglicherweise nicht wirksam oder sicher.
Beim Kauf legaler Steroide sind Qualität und
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legalen Steroiden. Dazu zählen zum Beispiel PayPal
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Mehrere Nahrungsergänzungsmittel können die Produktion von menschlichen Wachstumshormonen erhöhen. Eine ausreichende Menge an Tiefschlaf
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Kohlenhydrate und Zucker erhöhen den Insulinspiegel am meisten, so dass
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In Mexiko und bestimmten südostasiatischen Ländern kann HGH rezeptfrei gekauft werden.
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Oxandrolone Anavar, Oxandrin Reviews And User Ratings: Effectiveness, Ease Of Use, And Satisfaction
User‑Generated Reviews of Antidepressants
A Guide to Reading, Understanding, and Interpreting Online Feedback
—
1. Introduction – Why Look at Online Reviews?
Personal experience matters: Clinical trials often involve small samples and strict protocols; real‑world use can reveal
nuances that studies miss.
Diverse populations: Reviews frequently cover people of different ages, ethnicities, comorbid conditions, and lifestyles—information useful
for gauging how a drug might work in varied settings.
Rapid updates: New side‑effect reports or efficacy observations can appear quickly on review platforms before they reach formal publications.
2. Finding Credible Review Sources
Platform Typical Strengths Common Caveats
Healthgrades, RateMDs (for prescribers) Links prescriptions to clinician ratings Limited
to patients who use the platform; no clinical detail
Drugs.com “Reviews” User‑generated but often includes dosage/side‑effect details Potential for bias
from dissatisfied users
FDA’s MedWatch reports Official adverse event reporting Under‑reporting is common; not peer reviewed
Peer‑reviewed journal commentaries (e.g., Annals
of Pharmacotherapy) Scientific rigor Not real‑time; limited to experts
Takeaway: Cross‑check user reviews with official sources.
If a drug shows an alarming number of “very bad” ratings,
look for FDA adverse event reports or published studies discussing safety signals.
—
3. How do I compare costs and insurance coverage?
A. Use pharmacy benefit manager (PBM) tools
Tool What it offers
GoodRx Lists price variations by pharmacy, coupons, and generic alternatives.
PharmacyChecker Shows average cost for specific drug brands/generics.
Cigna/Blue Cross member portal Provides personalized formulary status and copay amounts.
B. Look at the patient’s formulary
Identify the drug’s tier (e.g., Tier 2 = moderate copay, Tier 3 = higher).
Check if there are preferred generics or alternate agents that provide equivalent therapy.
C. Consider insurance coverage and prior authorization
Prior Authorization: Some drugs require a pre‑approval from
the insurer; this can delay access.
Step‑Therapy: The insurer may insist on trying cheaper options first.
D. Evaluate patient financial assistance programs
Many manufacturers offer copay cards or patient assistance for qualifying patients,
which might reduce out‑of‑pocket costs dramatically.
—
3. Practical Decision‑Making Framework (Pseudocode)
Below is a high‑level algorithm you can implement
mentally or in your EMR to weigh the options quickly:
Input variables
disease_severity = severity_score
e.g., 0–10
clinical_evidence_strength = evidence_score
0–5 scale
side_effect_profile = side_effect_risk
0–3 scale
cost_out_of_pocket = cost_value
dollars/month
Weighted scoring
weight_severity = 4
weight_evidence = 3
weight_side_effects = 2
weight_cost = 1
score = (disease_severity weight_severity) + \
(clinical_evidence_strength weight_evidence) – \
(side_effect_profile weight_side_effects) – \
(cost_out_of_pocket / 1000 weight_cost)
Decision thresholds
if score >= 10:
recommendation = “Proceed with treatment”
elif score = 50 and monthlyCost diseaseSeverity >= 50?
Yes. Monthly cost = 50.
So why would it be not affordable? But the sample says
“Monthly cost exceeds affordability threshold.” So perhaps there
is a different meaning of ‘affordability threshold’ than 500.
Maybe 500 is not the affordability threshold but
something else.
Wait, let’s read again carefully.
In the problem description:
A user has a monthly budget for entertainment expenses and wants to
know if their planned activities are within this budget.
The user inputs:
Monthly budget amount (e.g., $200).
Number of days in the month (e.g., 30).
Number of days they plan to spend on entertainment activities.
Estimated cost per day for these activities.
Then, calculate total estimated entertainment cost: number_of_entertainment_days estimated_cost_per_day.
If total_estimated_entertainment_cost > monthly_budget:
Output a warning message stating that the planned
expenses exceed the user’s budget.
Else:
No output is required.
So, in effect, if the total estimated entertainment cost exceeds the
monthly budget, output a warning message; else,
no output.
Additionally, the constraints specify that all inputs are positive real
numbers. So we don’t need to handle negative or zero values.
Also, the example input and output match this logic:
Input:
1000
5
200
The calculation is 5 200 = 1000, which equals the monthly budget,
so no output is expected.
But in the problem description, it says “If the total estimated entertainment cost exceeds the user’s monthly budget” (i.e., strictly greater than).
So if it’s equal to the budget, then there’s no warning.
Wait, but in the example input and output:
Input: 1000
5
200
Output: (No output)
So calculation is 5200=1000, which equals the budget. No output expected.
Thus, our logic should be that if total estimated entertainment cost
> monthly budget => print warning.
But then later in “Your Task” section:
If the total estimated entertainment cost exceeds the user’s
monthly budget, print a warning message: “Warning: Your entertainment expenses exceed your monthly budget.”
But in the example, 1000=budget, no output.
So that matches.
Thus, we need to implement this logic.
Now, what about the constraints and subtasks?
The problem seems to be from some competitive programming site where they have different subtasks with increasing
constraints.
They define:
Subtask
(10 points): N = 2
Subtask
(20 points): 2 = 2)
Subtask
(40 points): No constraints.
But this is more relevant for performance considerations.
But given that we need to write a solution in JavaScript, we just
need to produce correct answer.
Also note: The input format uses N and M as numbers but then they give the next line contains N integers Ai, each between 1 and 1000.
Then second part: “M lines follow.” Each line starts with an integer
Mi (the number of items in the i-th box), followed by a list of Mi integers describing which types are present in that box.
But we need to interpret Mi as the count of items in the box,
but they also give each type of item. But careful:
They say “each line starts with an integer M_i (the number of items in the i-th box), followed by a list of M_i integers describing which types are present in that box.” So Mi
is just the number of items, not the number of distinct
types? Actually it’s likely each type appears once.
But maybe they can repeat? It’s not specified but probably each item has a type, and the list may contain duplicates if there are multiple items of same type.
But we don’t need to handle that nuance; we only need to know whether each box
contains at least one of each type in the set. We can treat duplicates as irrelevant because they count for presence anyway.
So the algorithm: For each i from 0..N-1, compute a bitmask mask_i such that mask_i
has bit t set if box i contains item of type t for any t in the
given set. Then we need to find minimal subset S of indices such that OR
over i∈S of mask_i equals fullMask (where fullMask = sum_t 1 boxes(m) // store
list of item prices.
for j in 0..m-1: read line; parse:
– istringstream ss(line);
– string boxName;
– int k;
– ss >> boxName >> k;
– for t=0..k-1: int p; ss >> p; boxesj.push_back(p); // store price.
Now compute dp array:
vector dp(10001, INF);
dp0 = 0;
For each item list in boxes:
// Optionally we can use copy of current dp
as prev
vector newdp = dp; // default same
for each price p in itemList:
for (int s=0; s dps) newdpns = dps;
// After processing all items for this box, update
dp
dp.swap(newdp);
// After all boxes processed:
// Find minimal total cost among sums >= M
int ansCost = INF;
for (int s=M; stotal sum of all boxes? But each box can be bought
at most once. Suppose there are N boxes; each has sum s_i.
Then maximum possible sum is sum_i s_i. We may not be able to reach
M if M > total sum. In that case, we cannot achieve target.
The problem statement says “you should get at least M” but doesn’t specify what happens if impossible.
But given constraints maybe always possible? Not necessarily.
But we can still compute; if not possible,
algorithm will return INF or some large number. We might output -1 or something?
But no spec. Maybe it’s guaranteed that solution exists.
We’ll assume solution exists.
Implementation details:
– We’ll read input lines via fs.readFileSync(0,’utf8′).trim().split(/\s+/).
But careful: if there is trailing newline, trim() will remove it; but we might lose last newline?
Not relevant.
We parse tokens sequentially. Since we need to know
when a test case ends, we can read N and M first, then for each of N lines, read M numbers.
That should be fine because tokens are contiguous. There is no sentinel value.
So we just iterate through tokens accordingly.
Edge Cases: The input may contain blank lines between tests;
splitting by whitespace will ignore them. Good.
Implementation Steps:
– const data = fs.readFileSync(0, ‘utf8’).trim().split(/\s+/).map(Number);
But careful: If there are trailing newlines after the last test
case, .trim() will remove them; fine.
Now parse:
let idx=0;
const T=dataidx++;
let outputs=;
for(let t=0;t new Array(nCols).fill(0));
for(let r=1; r best) best = maxHere;
if (best > overallBest) overallBest = best;
return overallBest;
But this is not efficient; we can optimize by computing colSums incrementally rather than recomputing each time.
Use loops accordingly.
Simpler: For each pair of top and bottom rows, accumulate column sums as
we iterate over bottom row. We’ll implement typical algorithm:
let best = -Infinity;
for (top=0; top +1 else -1;
Similarly colSum = new intN; for each column j, sum over i.
Also diag1Sum and diag2Sum: diag1 (top-left to
bottom-right) index i==j; diag2 (top-right to bottom-left) where i+j == N-1.
We can compute by loops:
for i=0..N-1:
for j=0..N-1:
int val = boardij.equals(“O”)? 1 : -1;
rowSumi += val;
colSumj += val;
if (i==j) diag1Sum += val;
if (i + j == N-1) diag2Sum += val;
Now we need to find winning line: rowSumi==N or rowSumi==-N etc.
We can check:
int winner = 0; // 1 for O, -1 for X
int winLineCount = 0;
int winnerLinesIndices; // maybe store them
We might just iterate again through each line to find winner and
also track number of winning lines.
Because we need to compute uniqueness: if more than one winning line
-> invalid. If exactly one winning line -> valid only if
board state corresponds to that line being completed last.
But we still must ensure there are no other lines with same color but
not considered? Actually if there’s a second winning line, it’s automatically invalid because game stops earlier;
can’t have two lines at the end of game. But what about scenario where there is one winning line but also another line of same color that could
exist due to previous moves? That would be
impossible because once you had that other line earlier, the game would have ended.
So if board has two winning lines (both color), it’s invalid regardless.
Thus algorithm:
Count wins for each color.
If both colors have at least one win: return 0.
If only one color has a win:
– Let winnerColor be that color; let otherColor be the opposite.
– Determine if there exists at least one winning line of
winnerColor. (Yes)
– Determine if there is any winning line of otherColor?
Shouldn’t happen because we already checked both colors none.
So no wins for otherColor.
– Check if board has at least one empty cell:
– If yes: return 1 (since game not finished, and last move
could have created win).
– If no: return 2 (board full, winner just won with last move;
since no more moves possible after that).
Else if there is no winning line for any color:
– Determine if board has empty cells:
– If yes: return 0 (game in progress)
– If no: ??? This scenario can’t happen because at least one win would exist if full.
But just to be safe, we can treat as 2? But not needed.
Now let’s test this logic with some examples:
Example1: board all X -> winner X, board full -> return 2 (makes sense).
Example2: board all O -> winner O, board full -> return 2.
Example3: Mixed but no wins and empty cells -> return 0.
Example4: Mixed with win for X maybe? Let’s test:
board:
“XXX”
“O.O”
“…”
X has a horizontal row top. Winner X. Board not full (some ‘.’), so return 1.
Now what about board:
“XXO”
“OOX”
“XOX”
Let’s check if any wins: rows: “XXO” none, “OOX” none,
“XOX” none; columns: col0 ‘X’,’O’,’X’ -> no; col1 ‘X’,’O’,’O’?
Actually row 1: O O X? Wait let’s fill: board:
Row0: X X O
Row1: O O X
Row2: X O X
Check rows: Row0 “XXO” not all same; Row1 “OOX” not; Row2 “XOX” no.
Columns: col0 X,O,X -> no; col1 X,O,O? Actually
row1 col1 is ‘O’, row2 col1 ‘O’ -> column 1 has X
O O -> not; col2 O X X -> not.
Diagonals: main diag X O X (positions (0,0),(1,1),(2,2)) = X,O,X
not all same; anti-diag O O X? positions (0,2)=’O’,(1,
1)=’O’,(2,0)=’X’ -> not.
So no winner: tie.
Thus output should be 3 lines with “tie” each line.
That is the correct answer. Let’s double-check problem statement:
“For each test case, write to standard output one of the following strings:” Actually they didn’t specify
per test case? But typical tasks: For each test case produce a string.
But here we have multiple games within a single test case (3 games).
So maybe for each game produce a string. But the problem says “For each test case” which could be ambiguous.
However, I think they intend to output one line per game as well because otherwise they’d say “for each game”.
Let’s search memory: There’s a known Kattis problem called “Tic-Tac-Toe” but it’s about computing final
board given some input; not this.
But the sample might show that for 3 games
we output 3 lines. Let’s assume that’s correct.
Thus expected output:
X
O
X
Each on its own line, no extra spaces.
Double-check: Could there be trailing newline after
last line? Usually yes but not shown.
Let’s produce final answer accordingly.X
O
X
References:
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