给出一张记录[tex=9.571x1.357]F/bby+cZhVWR3sISWFC7ol+G3JdBp+eO7GGb08Afm31IZtyuEteU5QEpN2Z3hfFc[/tex]，用FFT算法求出[tex=1.857x1.357]F/bby+cZhVWR3sISWFC7ou9kcNE90QEJsRNd1bxwD7A=[/tex]的离散谱[tex=2.143x1.357]s7MVnEg82cKHiPYGB9mntyVoBtllCd3XzVk6ZqBRMmM=[/tex].
举一反三
- set1 = {x for x in range(10) if x%2!=0} print(set1) 以上代码的运行结果为？ A: {1, 3, 5, 7, 9} B: {1, 3, 5, 7} C: {3, 5, 7, 9} D: {3, 5, 7}
- set1 = {x for x in range(10) if x%2!=0} set1.remove(1) print(set1) 以上代码的运行结果为？ A: {1, 3, 5, 7, 9} B: {1, 3, 5, 7} C: {3, 5, 7, 9} D: {3, 5, 7}
- 已知列表x=[3, 5, 6, 7, 9]，那么x[::-1]的结果是 A: [3, 9] B: [3, 5, 6, 7, 9] C: [3, 5, 6, 7] D: [9, 7, 6, 5, 3]
- 假设“☆”是一种新的运算，若3☆2=3×4，6☆3=6×7×8，x☆4=840(x＞0)，那么x等于： A: 2 B: 3 C: 4 D: 5 E: 6 F: 7 G: 8 H: 9
- 【计算题】5 ×8= 6×4= 7×7= 9×5= 2×3= 9 ×2= 8×9= 7×8= 5×5= 4×3= 5+8= 6 ×6= 3×7= 4×8= 9×3= 1 ×2= 9×9= 6×8= 8×0= 4×7=