I don’t know where the key is, but I suppose I ____________it at home.A. could have leftB. should have leftC. must leave
题目
I don’t know where the key is, but I suppose I ____________it at home.
A. could have left
B. should have left
C. must leave
相似考题
更多“I don’t know where the key is, but I suppose I ____________it at home.A. could have left ”相关问题
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第1题：
–– Host: Jack, come and sit in the sofa. Dinner will be ready in a minute. Could I get you something to drink?
–– Guest: ____.
A: No, don ’t trouble. I ’ve drunk enough
B: No, you couldn't. I'm not thirsty
C: Yes, you could. I ’d like some Coca cola
D: Yes, please. I'd like some Sprite
参考答案：D
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第2题：
____ I?know,?there?isn’t?such?a?word?in?English.
A、As much as
B、So far as
C、As long as
D、As soon as
参考答案：B
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第3题：
-- Ann is in hospital.
-- Oh, really? I __ know. I __ go and visit her.
A. didn’t; am going to B. don’t; would
C. don’t; will D. didn't; will
正确答案：D
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第4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;
}
}
正确答案：(1)alefti-1 (2)(left+right+1)/2 (3)while(leftright) (4)a[right)=a[i] (5)aleft i-1
(1)a，left，i-1 (2)(left+right+1)/2 (3)while(leftright) (4)a[right)=a[i] (5)a，left, i-1 解析：(1)a，left，i-1
递归排序左子区间，从left到i-1元素，不包括i元素。
(2)(left+right+1)/2
三者取中子程序median3(a，left，right)，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。
(3)while(leftright)
循环直到left和right相遇。
(4)a[right)=a[i]
若a[i]＞pivot则让a[right]=a[i]而让a[i]=pivot；。
(5)a，left, i-1
递归排序左子区间，从left到i-1元素，不包括i元素。 -
第5题：
I’d rather you ________ make any comment on the issue for the time being.
A) don’t B) wouldn’t C) didn’t D) shouldn’t
选C
would rather that……"宁愿……",that从句中用过去时表示现在或将来要做的事。 -
第6题：
分析下列程序的上界O和下界W。 for （i = 1; i ＜ n; i++) key= a[i]; int j=i-1 while（j＞=0 && a[j]＞key ) a[j+1]=a[j] j- - a[j+1]=key 该程序时间复杂度的上界是O（____)、下界是W（_____)。
n^4,n^4