估计下列各酸[tex=1.571x1.429]b9i2VgyUYFAd8vi4sLJJRAL3/evLyauVnDs1iVWin2M=[/tex]值及酸强度[tex=15.357x1.214]/EbUrFeazBtuRoUEXHPbAPoQEONDVlUNFDX0LJ8rce+ocEmvPjhqadli5z5jDbXwTXOiyr+6O1BeX0fuI+sMNA==[/tex]
举一反三
- set1 = {x for x in range(10)} print(set1) 以上代码的运行结果为？ A: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} C: {1, 2, 3, 4, 5, 6, 7, 8, 9} D: {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
- >>>x= [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9]>>>print(x.sort()) 语句运行结果正确的是( )。 A: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] B: [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9] C: [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] D: ['2', '4', '0', '6', '10', '7', '8', '3', '9', '1', '5']
- 以下程序段实现的输出是（）。for(i=0;i<;=9;i++)s[i]=i;for(i=9;i>;=0;i--)printf("%2d",s[i]);[/i][/i] A: 9 7 5 3 1 B: 1 3 5 7 9 C: 9 8 7 6 5 4 3 2 1 0 D: 0 1 2 3 4 5 6 7 8 9
- 下面说法错误的是（ ）。知识点：列表推导式 A: dict([(x, x**2) for x in range(6)]) 创建的字典是{0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25} B: [[x*3+y for y in range(1,4)] for x in range(3)] 创建的是二维列表 [[1, 2, 3], [4, 5, 6], [7, 8, 9]] C: number = [-2, 4, 6, -5]string = 'ab'z = [(i, j) if i>0 else (-i, j) for i in number for j in string]这段代码创建的列表为[(2, 'a'), (2, 'b'), (4, 'a'), (4, 'b'), (6, 'a'), (6, 'b'), (5, 'a'), (5, 'b')] D: ' '.join([i for i in range(1,11)])的运算结果为字符串'1 2 3 4 5 6 7 8 9 10'
- 【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=