问题 阅读理解

阅读理解。

      Most people in Britain live in small family groups. More than a quarter of homes in Britain have only got one

person in them. Some of these are old people but some are people of twenty or thirty who choose to live alone.

35% of homes have two people in them, and another 17% have three people. 15% have four people in them, and

the other homes have five or more. 

     The families in Britain are small. It is unusual for parents to have more than two children. When children are

about eighteen or nineteen, they leave their parents' homes, and they often go to different cities. Sometimes they

only visit their parents two or three times a year.

1. How many homes in Britain have 3 people living in them? [ ]

A. 8% of homes.

B. 15% of homes.

C. 17% of homes

D. 35% of homes.

2. At what age do children in Britain usually leave their parents' homes? [ ]

A. About 18 or 19.

B. About 20 or 21.

C. About 25 or 26.

D. About 30 or 31.

3. What is the passage? [ ]

A. An invention.

B. A report.

C. A. play

D. A joke.

4. What does the passage mainly tell us? [ ]

A. Young people in Britain.

B. Old people in Britain.

C. Families in Britain.

D. The population in Britain.

5. According to the passage, which of following is right? [ ]

A. Only once a year.

B. Only twice or three times a month.

C. Often.

D. Twice or three times a year.

答案

1-5 CABCD

多项选择题

【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;

问答题 简答题