The initial cost of a product and the product's operation and maintenance costs _____ .
A . increases perceived value when balanced.
B . are incidental to each other because initial costs are "sunk"
C . decreases design costs as operation periods increase
D . provides perceived function value and product social dysfunction
E . are integrally related with each other because initial costs are "sunk".
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)
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)
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)
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)
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; }
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)