Write the search algorithm for 89 in the following number sequence: 21.3.0.9.15.72.89.91.93

For I: = 1 to n do if (a [i] = 89) then begin place: = I; break; end; if the data is big, I will use nlogn + logn fast row + dichotomy with C + +, which is about sort (& A [1], & A [n + 1]); k = n / 2; L = 1; r = n; while (a [k]! = = 89) {if (a [k] > 89) {