Since the five loops cover $\sum \left(2(a_i + b_i) - 4\right) = 100$ cells, $\sum (a_i + b_i) = 60.$ Every loop has an even number of cells, so no odd-by-odd rectangle can be exactly filled by loops; and filling a rectangle whose shortest even side is $e$ requires at least $\frac{e}{2}$ loops, since peeling off an outermost loop shrinks that side by exactly $2$ while splitting a rectangle into smaller ones only adds up such requirements. Now consider the outermost loops of a partition (those whose rectangles lie inside no other loop's rectangle): their rectangles tile the $10 \times 10$ square. If outermost rectangle $R_i$ has shortest even side $e_i,$ it uses $n_i \ge \frac{e_i}{2}$ loops and covers at most $10\,e_i$ cells. Summing over the tiling, $100 \le \sum 10\,e_i \le 20 \sum n_i = 100,$ so equality holds throughout: each $R_i$ spans the full $10$ in one direction, has even width $e_i,$ and is filled with exactly $\frac{e_i}{2}$ loops. Two full-length slabs in different directions would overlap, so the outermost rectangles are the whole square or parallel slabs, and the same equality argument repeats inside every loop's inner rectangle.

Let $s(w)$ be the number of ways to fill a full-height slab of even width $w$ with $\frac{w}{2}$ loops, subject to the slab's own $10 \times w$ boundary being one outermost loop. (A split into smaller outermost slabs is counted later instead.) A width-$2$ slab is a single loop: $s(2) = 1.$ A width-$4$ slab is a $10 \times 4$ loop around an $8 \times 2$ loop: $s(4) = 1.$ A width-$6$ slab is a $10 \times 6$ loop around an $8 \times 4$ region holding two loops — either nested ($8 \times 4$ around $6 \times 2$) or two $8 \times 2$ slabs — so $s(6) = 2.$ A width-$8$ slab surrounds an $8 \times 6$ region holding three loops: an $8 \times 6$ loop around a $6 \times 4$ region with two loops ($2$ ways as before), or full-height strips of widths $2 + 2 + 2$ ($1$ way), or widths $2 + 4$ in two orders ($2$ ways), so $s(8) = 5.$ The same recursion counts the full square: a $10 \times 10$ loop around an $8 \times 8$ region with four loops, where the $4 \times 4,$ $6 \times 6,$ and $8 \times 8$ regions admit $3,$ then $3 + 3 + 3 = 9,$ then $9 + 9 + 9 = 27$ fillings (single nested loop, vertical strips, or horizontal strips at each stage).

Finally, tally the outermost structures. The single $10 \times 10$ rectangle gives $27$ partitions. For parallel slabs, the widths form a composition of $10$ into even parts with at least two parts, and orientations (vertical or horizontal) double the count: $(2,2,2,2,2)$ gives $1;$ $(4,2,2,2)$ in $4$ orders gives $4;$ $(4,4,2)$ in $3$ orders gives $3;$ $(6,2,2)$ in $3$ orders gives $3 \cdot 2 = 6;$ $(6,4)$ in $2$ orders gives $2 \cdot 2 = 4;$ and $(8,2)$ in $2$ orders gives $2 \cdot 5 = 10,$ for $28$ per orientation. The total is $27 + 2 \cdot 28 = 83.$