That is the existing product range. “product range” means product portfolion. ()此题为判断题(对,错)。

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That is the existing product range. “product range” means product portfolion. ()

此题为判断题(对,错)。


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  • 第1题:

    已知程序如下,该程序实现的功能为_____。 (10) main() (20) { int counter; (30) ... //输入N值的语句,假设N为偶数,略 (40) long product = 1; (50) for counter = 1 to N step 2 (60) { product = product * counter; } (70) return product; (80) }

    A.product = 1*3*5*...* (N-1)

    B.product = 1*2*3*...*(N-1)

    C.product = 1+ 2+3+...+ (N-1)

    D.product = 1+3+5+...+(N-1)


    sum = 1!+2!+...+n!

  • 第2题:

    已知程序如下,该程序实现的功能为_____。 (10) main() (20) { int counter; (30) ... //输入N值的语句,略 (40) long product = 1; (50) for counter = 1 to N step 2 (60) { product = product * counter; } (70) return product; (80) }

    A.product = 1*2*3*...*(N-1)

    B.product = 1+ 2+3+...+ (N-1)

    C.product = 1*3*5*...* (N-1)

    D.product = 1+3+5+...+(N-1)


    product = 1*3*5*...* (N-1)

  • 第3题:

    33、已知程序如下,该程序实现的功能为_____。 (10) main() (20) { int counter; (30) ... //输入N值的语句,略 (40) long product = 1; (50) for counter = 1 to N step 2 (60) { product = product * counter; } (70) return product; (80) }

    A.product = 1*2*3*...*(N-1)

    B.product = 1+ 2+3+...+ (N-1)

    C.product = 1*3*5*...* (N-1)

    D.product = 1+3+5+...+(N-1)


    product = 1*3*5*...* (N-1)

  • 第4题:

    41、已知程序如下,该程序实现的功能为_____。 (10) main() (20) { int counter; (30) ... //输入N值的语句,略 (40) long product = 1; (50) for counter = 1 to N step 2 (60) { product = product * counter; } (70) return product; (80) }

    A.product = 1*3*5*...* (N-1)

    B.product = 1*2*3*...*(N-1)

    C.product = 1+ 2+3+...+ (N-1)

    D.product = 1+3+5+...+(N-1)


    product = 1*3*5*...* (N-1)

  • 第5题:

    如下程序用来计算公式1!+2!+3!+…+10!,请完善程序。 #include <iostream> using namespace std; int main() { int i,j,sum,product; cout<<"1!+2!+3!+.......+10!= "; sum=【1】; for (【2】) { product=【3】; for (【4】) { product*=j; } sum+=product; } cout<<sum<<endl; return 0; }


    40

  • 第6题:

    6、已知程序如下,该程序实现的功能为_____。 (10) main() (20) { int counter; (30) ... //输入N值的语句,略 (40) long product = 1; (50) for counter = 1 to N step 2 (60) { product = product * counter; } (70) return product; (80) }

    A.product = 1*2*3*...*(N-1)

    B.product = 1+ 2+3+...+ (N-1)

    C.product = 1*3*5*...* (N-1)

    D.product = 1+3+5+...+(N-1)


    sum = 1!+2!+...+n!