题目详情
题目详情:
发布时间:2023-10-23 07:53:08

[单项选择]Project work packages are typically decomposed into smaller components called activities to provide a basis for (72), scheduling, executing, and monitoring and controlling the project work.
A. reviewing
B. estimating
C. auditing
D. expecting

更多"Project work packages are typically"的相关试题:

[单项选择]Project work packages are typically decomposed into smaller components called activities to provide a basis for (72) , scheduling, executing, and monitoring and controlling the project work.
A. reviewing
B. estimating
C. auditing
D. expecting
[单项选择]All technical work is completed on the project.Which of the following remains to be done ?()
A. Verify scop
B. Risk response pla
C. Staffing management pla
D. Lessons learne
[单项选择]A project management team has subcontracted work to a service company.Which of the following tools is the best choice to assure that this company will be able to meet the quality requirements of the project ?()
A. Quality audi
B. Deliverable inspectio
C. Fixed price contrac
D. Service level agreemen
[单项选择]The work of Project Manager is chiefly concerned with
A. emergency relief programmes.
B. agricultural rehabilitation.
C. helicopter assisted surveys.
D. strategic planning,
[单项选择]( )(Project Dragram,或称WBS Work Breakdown Structure)是一个组织工具,它通过树状图的方式对一个项目的结构进行逐层分解,以反映组成该项目的所有工作任务。
A. 组织结构图
B. 项目结构图
C. 工作任务分工表
D. 工作流程图
[单项选择]

有以下程序
int a=4;
int f(int n) {
int t=0;
static int a=5;
if(n%2) {int a=6;t+=a++;
}
else {
int a=7;t+=a++;
}
return t+a++;
}
main( ) {
int s=a,i=0;
for(;i<2;i++)s+=f(i);
printf("%d/n",s);
}
程序运行后的输出结果是()


A. 24
B. 28
C. 32
D. 36
[单项选择]有以下程序   void fun(int *a,int i,int j)   { int t;    if (i<j)    { t=a[i];a[i]=a[j];a[j]=t;    fun(a,++i,--j);    }   }   main( )   { int a[ ]={1,2,3,4,5,6},i;    fun(a,0,5);    for(i=0;i<6;i++)    printf("%d",a[i]);   }   执行后输出的结果是( )。
A. 6 5 4 3 2 1
B. 4 3 2 1 5 6
C. 4 5 6 1 2 3
D. 1 2 3 4 5 6
[单项选择]

有以下程序
  void f(int a[],int i,int j)
  { int t;
   if(i<j)
   { t=a[i]; a[i]=a[j];a[j]=t;
    f(a,i+1,j-1); } }
  main( )
  { int i,aa[5]={1,2,3,4,5};
   f(aa,0,4);
   for(i=0;i<5;i++) printf(“%d,”,aa[i]); printf(“/n”); }
   执行后输出结果是()


A. 5,4,3,2,1,
B. 5,2,3,4,1,
C. 1,2,3,4,5,
D. 1,2,3,4,5,
[单项选择]有以下程序: int fa(int x) { return x*x; } int fb(int x) { return x*x*x; } int f(int(*f1)( ),int(*f2)( ),int x) return f2(x)-f1(x); } main( ) { int i; i=f(fa,fb,2);printf("%d/n",i); } 程序运行后的输出结果是( )。
A. 1 2 6 8
B. 8 6 2 1
C. 8 1 6 2
D. 8 6 1 2
[单项选择]有以下程序 int fun(int x[],int n) { static int sum=0,i; for(i=0;iA. 45
B. 50
C. 60
D. 55
[单项选择]有如下程序段: int i=4;int j=l; int main( ){ int i=8,j=i; cout<<i<<j<<endl; } 运行时输出结果是
A. 44
B. 41
C. 88
D. 81
[单项选择]有以下程序 #include int fun(int a,int b) { if(b==0) return a; else return(fun(--a,--b)); } main( ) {printf("%d/n",fun(4,2));} 程序的运行结果是
A. 1
B. 2
C. 3
D. 4

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