(1) for i 1 to n-1 这个 (2) minval A[i] (3) minindex i (4) for j=ito n (5) if (A[j]<minval) 多 行 (6) minval A[j] (7) minindex =j (8) exchange A[i]and A[minindex] 次比 (9) bigjump =0 较操 (10) for i= 2 to n 作？ (11) if(A[i]>2*A[i-1]) (12) bigjump bigjump 1 easily solvable pieces.If we can decompose the problem into smaller pieces and so the smaller pieces,then we may be able to ethr add or multiply soitosmaller problemsinrder to solve the larger problem.n this这个 算法 需要 执行 多少 次比 较操 作？