If p is the perimeter of a right angle triangle with integral length sides, { a, b, c }, there are exactly three solutions for p = 120.
{[20,48,52], [24,45,51], [30,40,50]}
Given any value of p, return a cell array whose elements are the sorted side lengths of a possible right triangle whose perimeter is p. Furthermore, the elements of the output should be sorted by their shortest side length.
Solution Stats
Problem Comments
6 Comments
Solution Comments
Show comments
Loading...
Problem Recent Solvers2035
Suggested Problems
-
Back to basics 6 - Column Vector
1102 Solvers
-
Sum of diagonal of a square matrix
1633 Solvers
-
11533 Solvers
-
Remove entire row and column in the matrix containing the input values
551 Solvers
-
Relative ratio of "1" in binary number
1592 Solvers
More from this Author56
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Hi Team,
I am getting error after submitting as,
"We're sorry, but something went wrong. We've been notified about this issue and we'll take a look at it shortly"
But completion bar has been incremented.
Thanks and Have a Good DaY
Well I don't know why I can't pass the test
could anyone help
function c = right_triangle_sides(p)
c = {};
k=1;
for i=1:ceil(p/3)
for j=i:ceil((p-i)/2)
if i^2+j^2==(p-i-j)^2
c(k)=mat2cell([i j p-i-i],1);
k=k+1;
end
end
end
end
To creator of this group,
Please check the test case 3 in the Problem of calculating side length from area of triangle. One ';' is missing which is not allowing the case to pass for correct solution.
This problem really highlights problems with the Cody scoring system. The score using a double for loop to brute force the lengths beats out a single for loop using equations. I tested my code in Matlab vs double loops and it's faster than most double loops I tested by several orders of magnitude for larger numbers (p>1000). There is also some sort of caching going on where my code gets orders of magnitude faster again if ran a few times vs most double for loops.
There is one exception with a double loop that is really fast by Hung Hoang. They use the for loops in a way that doesn't make it O^2.
interesting problem
đánh giá là bài lỗi vãi đái:))