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【简答题】
阅读理解。 Life is like throwing and catching four balls of life in the air．You call the balls work,family,friends and minds．You will soon understand that work is a rubber ball．If you drop it,it will bounce back (弹回)．The other three balls are made of glass and easy to be bro- ken,so you must know how to protect them．But how? Don't shut love out of your family by saying it's hard to find．The quickest way to get love is to give it; the fastest way to lose love is to hold it closely; and the best way to keep love is to give it wings． Don't think that you are not so good as others．It is because we are different and each of us is special．Never mind if you fall behind others．It's never too late to learn．Knowledge is a weightless treasure which you always carry easily． Don't let you life pass through your fingers by living in the past or in the future．If you live well today, you live all the days of your life． Yesterday is history,tomorrow is hard to guess and today is gift．That's why we call it 'The Present'． 1．Which is not the name of a ball talked about in the passage? A．work B．health C．family D．mind 2．To give it wings means to______ A．hold it closely B．give love freedom C．get ready for love D．keep love out of your family 3．Why is knowledge a treasure you can always carry easily? A．Because it is heavy． B．Because you can't protect it． C．Because it's never too late to learn． D．Because it is treasure without weight． 4．From the fourth paragraph,we know that A．people must have plans for the future B．people woul d rather live in the past C．people who live well today will live well every day． D．people who live in the past or in the future will not have long life．
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【单选题】下列算法实现在顺序表 L 的第 i ( 1 ≤ i ≤ L->length+1 )个结点的位置上插入值为 t 的元素,其中 listsize成员 为顺序表 L 的容量,表中第 1 个结点的数据存放在数组元素 L->elem[0] 中。下列选项中能完成此功能的语句序列为( )。 void InsertList(SeqList *L ,int i ,ElemType e) {// 将 e 插入顺序表 ...
A.
①L->elem[j-1]=L->elem[j]; ②L->elem[i]=e; ③L->length+1;
B.
①L->elem[j-1]=L->elem[j]; ②L->elem[i-1]=e; ③L->length++;
C.
①L->elem[j+1]=L->elem[j]; ②L->elem[i-1]=e; ③L->length++;
D.
①L->elem[j+1]=L->elem[j];②L->elem[i]=e; ③L->length+1;
【单选题】下列算法实现在顺序表 L 的第 i ( 1 ≤ i ≤ L->length+1 )个结点的位置上插入值为 t 的元素,其中 ListSize 为顺序表 L 的容量,表中第 1 个结点的数据存放在数组元素 L->data[0] 中。下列选项中能完成此功能的语句序列为( )。 void InsertList(SeqList *L,DataType t,int i) {// 将 t 插入顺序表 L 的第...
A.
①L->data[j-1]=L->data[j]; ②L->data[i]=t; ③L->length+1;
B.
①L->data[j-1]=L->data[j]; ②L->data[i-1]=t; ③L->length++;
C.
①L->data[j+1]=L->data[j]; ②L->data[i-1]=t; ③L->length++;
D.
①L->data[j+1]=L->data[j];②L->data[i]=t; ③L->length+1;
【简答题】插入算法: int InsertList(SeqList *L, DataType t, int i) {// 将 t 插入顺序表 L 的第 i 个结点的位置上 int j; if ( 1. ) { puts(" 插入位置错 "); return 0; } if ( 2. ) { puts(" 表满不能插入 "); return 0; } for( 3. ) 4. ; // 结点依次后移 L->d...
【单选题】下列算法实现在顺序表 L 的第 i ( 1 ≤ i ≤ L->length+1 )个结点的位置上插入值为 t 的元素,其中 ListSize 为顺序表 L 的容量,表中第 1 个结点的数据存放在数组元素 L->data[0] 中。下列选项中能完成此功能的语句序列为( )。 int InsertList(SeqList *L,DataType t, int i) {// 将 t 插入顺序表 L 的第...
A.
① L->data[j-1]=L->data[j]; ② L->data[i]=t; ③ L->length+1;
B.
① L->data[j-1]=L->data[j]; ② L->data[i-1]=t; ③ L->length++;
C.
① L->data[j+1]=L->data[j]; ② L->data[i-1]=t; ③ L->length++;
D.
① L->data[j+1]=L->data[j]; ② L->data[i]=t; ③ L->length+1;
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