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E. Nauuo and Cards (Simplified Version)
Nauuo has 2n cards:
n real cards numbered from 1 to n
n empty cards, represented as 0
These cards are randomly shuffled and split into:
Nauuo’s hand → a list of n cards
The pile → another list of n cards, ordered top to bottom
✅ Operation
She can:
Choose any card from her hand, and
Play it — move it to the bottom of the pile,
Then draw the top card of the pile into her hand.
She wants the pile to end up as [1, 2, 3, ..., n] (from top to bottom) as fast as possible.
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Answer :
```
def main():
def iinp(): return (int(input()))
def linp(): return (list(map(int, input().split())))
n = iinp()
arr = [0 for i in range(n+1)]
for i in range(n-1):
u,v = linp()
arr[u] += 1
arr[v] += 1
for i in arr:
if i ==2:
print("NO")
return
print("YES")
main()
```\
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▎🧩 D. Add on a Tree (Simplified)
You are given a tree (a connected graph with no cycles) of n nodes. Each edge in the tree initially has a value of 0.
You can perform the following operation:
> Choose any two leaf nodes (nodes connected to only one other node), and a real number x.
> Then, add
x to all edges on the simple path between these two leaf nodes.
🔁 You can repeat this operation as many times as you want, with different pairs of leaves and values.
---
▎❓ Question
Is it possible to reach any possible configuration of real numbers on the edges using a finite number of such operations?
Print:
• YES — if it’s always possible for this tree
• NO — if there exists any configuration that you cannot reach
---
▎📥 Input
• First line: n — number of nodes (2 ≤ n ≤ 10⁵)
• Next n-1 lines: two integers u and v, meaning an edge between node u and node v
It is guaranteed that the graph is a tree.
---
▎📤 Output
• Print YES or NO
---
▎📌 Examples
Input
2 1 2Output
YES--- Input
3 1 2 2 3Output
NO
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▎C. Labs (Simplified)
You have n² labs numbered from 1 (lowest) to n² (highest).
Any lab u can send 1 unit of water to any lower lab v if u > v .
▎Problem Statement
Split the labs into n groups of size n .
For each ordered pair of groups (A, B) , let f(A, B) be the total units sendable from all labs in A to all in B .
Objective: Maximize the minimum f(A, B) over all A ≠ B .
▎Output
Print any grouping that achieves this.
▎Example
Input:
3Output (one possible grouping):
2 8 5 9 3 4 7 6 1Here, n = 3 , meaning there are labs numbered from 1 to 9. Every group has 3 labs. The smallest f(A, B) among all 6 ordered pairs is 4, which is optimal. Solution
ddef main():
n = int(input())
groups = [[] for _ in range(n)]
num = 1
for row in range(n):
# decide direction: left‑to‑right on even rows, right‑to‑left on odd
cols = range(n) if row % 2 == 0 else range(n - 1, -1, -1)
for col in cols:
groups[col].append(num)
num += 1
# output
for g in groups:
print(*g)
if __name__ == "__main__":
main()1 261
▎B. Integer Points
▎Problem (brief)
DLS draws
n lines of the form y = x + pᵢ and JLS draws m lines of the form y = -x + qⱼ . Count how many pairs (one from each set) intersect at integer coordinates (x, y).
▎Key Fact
The intersection of y = x + p and y = -x + q is given by:
x = q - p / 2, y = q + p / 2For both x and y to be integers, q and p must have the same parity. ▎Solution 1. Count the number of even and odd integers in sets P and Q. 2. The total number of intersecting pairs is calculated as:
Answer = (even_P × even_Q) + (odd_P × odd_Q)▎Implementation
for _ in range(int(input())):
n, P = int(input()), list(map(int, input().split()))
m, Q = int(input()), list(map(int, input().split()))
eP = sum(p % 2 == 0 for p in P) # Count evens in P
oP = n - eP # Count odds in P
eQ = sum(q % 2 == 0 for q in Q) # Count evens in Q
oQ = m - eQ # Count odds in Q
print(eP * eQ + oP * oQ)
3 3 1 3 2 2 0 3 1 1 1 1 1 2 1 1▎Output
3 1 0
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here is the first question of A2SV weekly contest For G6 i will post each question with their answer
▎🖊 A. Pens and Pencils
▎Problem Statement
Tomorrow is a difficult day for Polycarp: he has to attend lectures and practical classes at the university! He writes lectures with pens and practicals with pencils.
• One pen lasts for
c lectures.
• One pencil lasts for d practicals.
• His pencil case can hold at most k writing tools in total.
Can Polycarp pack enough pens and pencils to cover the day?
---
▎Input Format
• First line: An integer t (1 ≤ t ≤ 100), the number of test cases.
• Each test case: Five integers a, b, c, d, k
– a: Number of lectures
– b: Number of practical classes
– c: Lectures per pen
– d: Practicals per pencil
– k: Maximum tools in the pencil case
---
▎Output Format
For each test case, output:
• Two integers x y, where:
– x: Number of pens
– y: Number of pencils
• Or output -1 if it’s not possible to pack enough tools.
---
▎Example Input
3 7 5 4 5 8 7 5 4 5 2 20 53 45 26 4▎Example Output
2 1 -1 1 3--- ▎Explanation • Test Case 1: – Needs
ceil(7/4) = 2 pens and ceil(5/5) = 1 pencil.
– Total = 2 + 1 = 3 ≤ 8 (possible).
• Test Case 2:
– Needs 2 pens and 1 pencil.
– Total = 2 + 1 = 3 > 2 (not possible).
• Test Case 3:
– Needs 1 pen (ceil(20/45)) and 3 pencils (ceil(53/26)).
– Total = 1 + 3 = 4 = 4 (possible).
---
▎Sample Code (Python)
t = int(input())
for _ in range(t):
a, b, c, d, k = map(int, input().split())
pens_needed = (a + c - 1) // c # Ceiling of a/c
pencils_needed = (b + d - 1) // d # Ceiling of b/d
if pens_needed + pencils_needed <= k:
print(pens_needed, pencils_needed)
else:
print(-1)
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if u have better ideas, i would love to hear and collaborate with u .please feel free to contact me with this bot @zprogramming_bot
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i want to make this channel more usefull for u so .what type of post do u want to see more often?
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Repost from N/a
Built something cool! 🚀
Verify is my personal project—an API for free, unlimited phone verification using Telegram. No SMS costs, simple integration, and optional self-hosting.
🔗 Check it out: https://verify.yonathan.tech
Would love your feedback!
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Hey Developers! Let me introduce you to an amazing project: an unlimited free phone verification API created by my friend Yonatan. Check it out!
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can u solve this question? the logic is simple but it may take while to figure it out
70. Climbing Stairs
Solved
Easy
You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Example 1:
Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: n = 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
Constraints:
1 <= n <= 45
see the question in leetcode
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Check this website to practice for the INSA test. There are some rumors that INSA takes questions from this site.
Mensa IQ Test
Leetcode with dani
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