• 2022-06-06
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    举一反三

    内容

    • 0

      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}

    • 1

      输出九九乘法表。 1*1=1 2*1=2 2*2=4 3*1=3 3*2=6 3*3=9 4*1=4 4*2=8 4*3=12 4*4=16 5*1=5 5*2=10 5*3=15 5*4=20 5*5=25 6*1=6 6*2=12 6*3=18 6*4=24 6*5=30 6*6=36 7*1=7 7*2=14 7*3=21 7*4=28 7*5=35 7*6=42 7*7=49 8*1=8 8*2=16 8*3=24 8*4=32 8*5=40 8*6=48 8*7=56 8*8=64 9*1=9

    • 2

      A=[1 2 3 4 5 6 7 8 9]A(5)=[]A=1 4 7 5 8 3 6 9

    • 3

      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

    • 4

      写出以下表达式的结果,结果之间间隔一个空格: 6 + 5 / 4 - 2 2 + 2 * (2 * 2 - 2) % 2 / 3 10 + 9 * ((8 + 7) % 6) + 5 * 4 % 3 * 2 + 3 1 + 2 + (3 + 4) * ((5 * 6 % 7 / 8) - 9) * 10