Going back from cumsum for a matrix

Suppose I cumsum a matrix, say A=rand(3,3); B=cumsum(A).
Knowing B, how to I get back to A, in a reasonably efficient way, for a rather large B?
Thanks!

 采纳的回答

A=round(rand(100,100),4);
B=cumsum(A);
a=round([B(1,:);diff(B)],4);
isequal(A,a)
ans = logical
1

1 个评论

Right.
Key points here are the use of diff(), the duplication of the first entry, and the rounding or other way of comparing with tolerance for the cross-check (since you would need to deal with round-off errors.)

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