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单选题
What are minimum and maximum number of extent pools on an IBM System Storage DS8800 with 64 physical disks（）
A
the minimum is 1 and the maximum is 8
B
the minimum is 2 and the maximum is 8
C
the minimum is 2 and the maximum is 16
D
the minimum is 1 and the maximum is 64
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更多 “单选题What are minimum and maximum number of extent pools on an IBM System Storage DS8800 with 64 physical disks（）A the minimum is 1 and the maximum is 8B the minimum is 2 and the maximum is 8C the minimum is 2 and the maximum is 16D the minimum is 1 and the maximum is 64” 相关考题
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阅读以下说明和C函数，填充函数中的空缺，将解答填入答题纸的对应栏内。
[说明]
下面的函数sort(int n,int a[])对保存在数组a中的整数序列进行非递减排序。由于该序列中的元素在一定范围内重复取值，因此排序方法是先计算出每个元素出现的次数并记录在数组b中，再从小到大顺序地排列各元素即可得到一个非递减有序序列。例如，对于序列6，5，6，9，6，4，8，6，5，其元素在整数区间[4，9]内取值，因此使数组元素b[0]～b[5]的下标0～5分别对应数值4～9，顺序地扫描序列的每一个元素并累计其出现的次数，即将4的个数记入b[0]，5的个数记入b[1]，依此类推，9的个数记入b[5]。最后依次判断数组b的每个元素值，并将相应个数的数值顺序地写入结果序列即可。
对于上例，所得数组b的各个元素值如下：
1.jpg
那么在输出序列中写入1个4、2个5、4个6、1个8、1个9，即得4，5，5，6，6，6，6，8，9，从而完成排序处理。
[C函数] void sort(int n,int a[]) { int *b; int i, k, number; int minimum=a[0],maximum=a[0]; /*minimum和maximum分别表示数组a的最小、最大元素值*/ for(i=1; i＜n; i++){ if(______) minimum=a[i]; eiSe if (______) maximum=a[i]; } number=maximum-minimum+1; if(number＜=i)return; b=(int*)calloc(number,sizeof(int)); if(!b) return; for(i=0;i＜n; i++){/*计算数组a的每个元素值出现的次数并记入数组b */ k=a[i]-minimum; ++b[k]; } /*按次序在数组a中写入排好的序列*/ i=______; for(k=0; k＜number; k++) for(; ______; --b[k] ) a[i++]=minimum+______; }
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单选题What are minimum and maximum number of extent pools on an IBM System Storage DS8800 with 64 physical disks（）A
the minimum is 1 and the maximum is 8B
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