統計
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解決済み
Swap between columns
The idea is to swap between second and second last column Ex = [1 2 3 4 5; 1 2 3 4 5; 1 2 3 4 5; 1 2 3 4 5; ...
7日 前
解決済み
Remove a specific character with another
Remove any (-) dash sign with (_) underscore Ex = 'The-Journey-of-thoudsands-miles-starts-with-a-single-step' y = 'The_Jour...
7日 前
解決済み
Swap between first and last
The idea is to swap between first and last row Ex = [1 2 3 4 5; 1 2 3 4 5; 1 2 3 4 5; 1 2 3 4 5; ...
7日 前
解決済み
Swap between first and last column
The idea is to swap between first and last column Ex = [1 2 3 4 5; 1 2 3 4 5; 1 2 3 4 5; 1 2 3 4 5; 1 2 3 ...
7日 前
解決済み
Swap between rows
The idea is to swap between second and second last row Ex = [1 2 3 4 5; 5 4 3 2 1; 1 2 3 4 5; 1 2 3 4 5; ...
7日 前
解決済み
Sum of series IX
What is the sum of the following sequence: Σ 1/k! for k=1...n for different n?
7日 前
解決済み
Sum of series
a(n) = n^2 - (n-1)^2 find the summation of the series upto n i.e. a(1)+a(2)+...+a(n)
7日 前
解決済み
Sum of series VI
What is the sum of the following sequence: Σk⋅k! for k=1...n for different n?
7日 前
解決済み
Sum of series V
What is the sum of the following sequence: Σk(k+1) for k=1...n for different n?
7日 前
解決済み
Sum of series IV
What is the sum of the following sequence: Σ(-1)^(k+1) (2k-1)^2 for k=1...n for different n?
7日 前
解決済み
Sum of series III
What is the sum of the following sequence: Σ(2k-1)^3 for k=1...n for different n?
7日 前
解決済み
Sum of series II
What is the sum of the following sequence: Σ(2k-1)^2 for k=1...n for different n?
7日 前
解決済み
Sum of series I
What is the sum of the following sequence: Σ(2k-1) for k=1...n for different n?
7日 前
解決済み
Sum of series VII
What is the sum of the following sequence: Σ(km^k)/(k+m)! for k=1...n for different n and m?
7日 前
解決済み
The MATLAB Treasure Hunt – Escape the Labyrinth of Logic by Unlocking the Correct Door Sequence
You enter the Labyrinth of Logic, where each door opens only if a secret rule is met. A vector of integers represents door code...
10日 前
解決済み
The MATLAB Treasure Hunt – Locate the Hidden Treasure in the Chamber of Coordinates
Inside the Chamber of Coordinates, glowing runes show several coordinate points on a 2D grid. The treasure lies closest to the ...
10日 前
解決済み
The MATLAB Treasure Hunt – Cross the River of Ratios by Finding Successive Proportions in the Data Stream
Following the glowing script, you arrive at the River of Ratios — a flowing stream of numbers. A carved message on the rocks re...
10日 前
解決済み
The MATLAB Treasure Hunt – Decode the Ancient Script by Removing Strange Symbols from the Message
Emerging from the Maze of Numbers, you find a tablet etched with symbols and letters — an Ancient Script! The message is hidden...
10日 前
解決済み
The MATLAB Treasure Hunt – Identify the Hidden Pattern Within the Maze of Numbers
The compass leads you to an underground chamber covered in numbers carved into the walls — a maze of digits that glows in myster...
10日 前
解決済み
The MATLAB Treasure Hunt – Fix the Broken Compass by Normalizing Its Angle Readings
Your compass, recovered from the Hidden Cave, spins wildly! The ancient device shows angles that exceed 360° or drop below 0°. T...
10日 前
解決済み
The MATLAB Treasure Hunt – Count Matching Symbols on the Ancient Scroll to Unlock the Next Clue
Inside the Hidden Cave, you discover an ancient scroll filled with strange symbols and markings. A note beside it says: “Count ...
10日 前
解決済み
The MATLAB Treasure Hunt – Extract the Hidden Cave’s Coordinates from a Long List of Numbers
After decoding the first clue, you uncover another piece of parchment — a long list of numbers. At first glance, it looks random...
10日 前
解決済み
The MATLAB Treasure Hunt – Decode the First Clue Hidden in a Jumbled Sequence of Numbers
You discover an ancient parchment inside the college archives. It contains a jumbled sequence of numbers that seems meaningless ...
10日 前
解決済み
Given a square and a circle, please decide whether the square covers more area.
You know the side of a square and the diameter of a circle, please decide whether the square covers more area.
11日 前
解決済み
Approximation of Pi
Pi (divided by 4) can be approximated by the following infinite series: pi/4 = 1 - 1/3 + 1/5 - 1/7 + ... For a given number of...
11日 前
