According to Cody Problem 1516, if I put 3 balls into 2 boxes then
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
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.
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