A. cheerful
B. for sure
C. at ease
D. in advance
第1题:
A、oughtn't to leave
B、needn't have left
C、needn’t leave
D、couldn’t leave
第2题:
A.ask
B.made
C.let
D.had
第3题:
阅读下列C程序和程序说明,将应填入(n)处的字句写在对应栏内。
【说明】下面是一个用C编写的快速排序算法。为了避免最坏情况,取基准记录pivot时,采用从left、right和mid=[(left+right)/2]中取中间值,并交换到right位置的办法。数组a存放待排序的一组记录,数据类型为T,left和right是待排序子区间的最左端点和最右端点。
void quicksort (int a[], int left, int right) {
int temp;
if (left<right) {
hat pivot = median3 (a, left, right); //三者取中子程序
int i = left, j = right-1;
for(;;){
while (i <j && a[i] < pivot) i++;
while (i <j && pivot < a[j]) j--;
if(i<j){
temp = a[i]; a[j] = a[i]; a[i] = temp;
i++; j--;
}
else break;
}
if (a[i] > pivot)
{temp = a[i]; a[i] = a[right]; a[right] = temp;}
quicksort( (1) ); //递归排序左子区间
quieksort(a,i+1 ,right); //递归排序右子区间
}
}
void median3 (int a[], int left, int right)
{ int mid=(2);
int k = left;
if(a[mid] < a[k])k = mid;
if(a[high] < a[k]) k = high; //选最小记录
int temp = a[k]; a[k] = a[left]; a[left] = temp; //最小者交换到 left
if(a[mid] < a[right])
{temp=a[mid]; a[mid]=a[right]; a[right]=temp;}
}
消去第二个递归调用 quicksort (a,i+1,right)。 采用循环的办法:
void quicksort (int a[], int left, int right) {
int temp; int i,j;
(3) {
int pivot = median3(a, left, right); //三者取中子程序
i = left; j = righi-1;
for (;; ){
while (i<j && a[i] < pivot)i++;
while (i<j && pivot <a[j]) j--;
if(i <j) {
temp = a[i]; a[j]; = a[i]; a[i]=temp;
i++; j--;
}
else break;
}
if(a[i]>pivot){(4);a[i]=pivot;}
quicksoft ((5)); //递归排序左子区间
left = i+1;
}
}
第4题:
A.don't
B.can't
C.couldn't
D.didn't
第5题:
He__me he was leaving on Wednesday.
A. said
B. told
C. tell
第6题:
– Do you agree with me? -- ______________.
A、No.
B、I’m afraid not.
C、I can’t.
D、Not.