1 1 1
1 1 2
1 2 2
2 2 2
The columns stand for a ball, the number stands for a box, and the row stands for a case.
Now we count the number of balls in each boxes.
so we get
3 0
2 1
1 2
0 3
the 1-th row stand for the 1-th case,3 balls are put into the first boxes,the others is empty the 2-th row stand for the 2-th case,2 balls are put into the first boxes,1 balls is put into the second boxes
You should write a function, input m and n, m stand for the number of balls ,n stand for the number of boxes,output all the cases.
Example
If m = 3,n = 4, you should output
3 0 0 0
2 1 0 0
2 0 1 0
2 0 0 1
1 2 0 0
1 1 1 0
1 1 0 1
1 0 2 0
1 0 1 1
1 0 0 2
0 3 0 0
0 2 1 0
0 2 0 1
0 1 2 0
0 1 1 1
0 1 0 2
0 0 3 0
0 0 2 1
0 0 1 2
0 0 0 3
Solution Stats
Problem Comments
2 Comments
Solution Comments
Show comments
Loading...
Problem Recent Solvers63
Suggested Problems
-
Find third Side of a right triangle given hypotenuse and a side. No * - or other functions allowed
249 Solvers
-
Square Digits Number Chain Terminal Value (Inspired by Project Euler Problem 92)
253 Solvers
-
17590 Solvers
-
776 Solvers
-
Mysterious digits operation (easy)
315 Solvers
More from this Author17
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
I'm not sure what is the point of eliminating 'ifs'? At any rate they can be easily replaced by e.g. 'while' loops, so in my opinion you should either eliminate both of them or none.
Again, the best way to generate such a sequence is using depth-first search or recursion. By prohibiting the usage of IFs, all solutions are inefficient (because we can't cut the tree, and are forced to follow all branches).