Euclidean floor-sum transformation

For integers $n\ge 0$, $m>0$, write $F(n,a,b,m)={\sum }_{i=0}^{n-1}\lfloor (ai+b)/m\rfloor$. Euclidean division gives $a=qm+r$ and $b=sm+t$, with $0\le r,t<m$, henceFor $0<a<m$ and $0\le b<m$, put $y=an+b$. Interchanging the two axes in the remaining lattice point problem givesIf $y<m$, the original sum is zero; if $a=0$ after reduction, it is also zero. Each swap lowers the modulus as in the Euclidean algorithm, enabling logarithmically many reduction stages. For example $F(5,3,1,7)=3$, and the swapped sum $F(2,7,2,3)=0+3$ gives the same answer.