Use floor modulo for % in generated C

[sueszli]

@proc
def foo(dst: f32[5], i: index):
    dst[i % 5] = 1.0

Actual:

dst[i % 5] = 1.0f;

Called with i = -1 this writes dst[-1]. ASan reports
stack-buffer-underflow, WRITE of size 4.

Expected:

dst[exo_floor_mod(i, 5)] = 1.0f;

Bounds checking treats % as floor modulo, so it proves that i % 5 lies in
[0, 5). C truncates toward zero instead. Division already routes through
exo_floor_div for the same reason. Modulo did not.

[codecov]

Codecov Report

✅ All modified and coverable lines are covered by tests.
✅ Project coverage is 88.56%. Comparing base (defe172) to head (7ec9909).

Additional details and impacted files

@@           Coverage Diff           @@
##             main     #852   +/-   ##
=======================================
  Coverage   88.56%   88.56%           
=======================================
  Files         100      100           
  Lines       21917    21926    +9     
=======================================
+ Hits        19410    19419    +9     
  Misses       2507     2507           

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