Searched refs:BenchmarkLCumSum1000 (Results 1 – 2 of 2) sorted by last modified time
240 func BenchmarkLCumSum1000(t *testing.B) { benchCumSum(naiveCumSum, 1000, t) } func
238 func BenchmarkLCumSum1000(t *testing.B) { benchCumSum(naiveCumSum, 1000, t) } func