网友您好, 请在下方输入框内输入要搜索的题目:

题目内容 (请给出正确答案)

It takes two weeks for Smith's left hand to get entirely ( ).

A. cured

B. dedicated

C. healed

D. mended


参考答案

更多 “ It takes two weeks for Smith's left hand to get entirely ( ).A. curedB. dedicatedC. healedD. mended ” 相关考题
考题 有以下程序 main( ) { char *s[ ]={"one","two","three"},*p; p=s[1]; printf("%c,%s\n",*(p+1),s[0]); } 执行后输出结果是A.n,twoB.t,oneC.w,oneD.o,two

考题 有以下程序: matin() { char * s[]={"one", "two", "three"}, *p; p=s[1]; printf("% c, % s\n", *(p+1),s [o]); } 执行后的输出结果是______。A.n, twoB.t, oneC.w, oneD.o, two

考题 有以下程序: main() { char *S[]={"one","two","three"},*p; p=s[1]; printf("%c,%s\n", *(p+1), s[0]); } 执行后输出结果是( )。A.n,twoB.t,oneC.w,oneD.o,two

考题 About ________ of the workers in the factory were born in the ________.A.two-thirds, 1970 B.two-thirds, 1970s C.two-third,1970 D.two-third, 1970s

考题 12. About_______ of the workers in the factory were born in the __________.A. two- thirds;1970B. two- thirds ; 1970sC. two-third ; 1970D. two-third ; 1970s

考题 Hearing problems may be alleviated by changes in diet and exercise habits.A:removed B:worsened C:relieved D:cured

考题 设指针变量p指向双向链表中节点A,指针变量s指向被插入的节点X,则在节点A的后面插入节点X的操作序列为()A.p->right=s;s->left=p;p->right->left=s;s->right=p->right; B.p->right=s;p->right->left=s;s->left=p;s->right=p->right; C.s->left=p;s->right=p->right;p->right=s;p->right->left=s; D.s->left=p;s->right=p->right;p->right->left=s;p->right=s;

考题 2、在循环双链表的p所指结点之后插入s所指结点的操作是()A.p->right=s;s->left=p;p->right->left=s;s->right=p->right;B.p->right=s;p->right->left=s;s->left=p;s->right=p->right;C.s->left=p;s->right=p->right;p->right=s;p->right->left=s;D.s->left=p;s->right=p->right; p->right->left=s;p->right=s;

考题 设指针变量p指向双向链表中结点A,指针变量s指向被插入的结点X,则在结点A的后面插入结点X的操作序列为()A.p->right=s; s->left=p; p->right->left=s; s->right=p->right;B.s->left=p;s->right=p->right;p->right=s; p->right->left=s;C.p->right=s; p->right->left=s; s->left=p; s->right=p->right;D.s->left=p;s->right=p->right;p->right->left=s; p->right=s;

考题 在循环双链表的p所指节点之后插入s所指节点的操作是A.p->right=s;s->left=p;p->right->left=s;s->right=p->right;B.s->left=p;s->right=p->right;p->right->left=s;p->right=s;C.s->left=p;s->right=p->right;p->right=s;p->right->left=s;D.p->right=s;p->right->left=s;s->left=p;s->right=p->right;