题目内容

下列哪些动物是新冠病毒的常见宿主( )

A. 蝙蝠
B. 獾
C. 果子狸
D. 蚊子

查看答案
更多问题

顺序栈的入栈操作Seqstack *push_seqstack(Sqstack *s,elementtype x){if((1)){printf(“溢出\n”);return 0;}else{s->top++;(2);return s;}}

链栈的出栈操作int *pop_linkstack(linkstack *s,elementtype y){linkstack *p;if((1)){printf(“溢出\n”);return 0;}else{p=s;y=s->data;s= (2);free(p);return 1;}}

循环队列的出队操作elementtype delete_cyqueue(Cqueue *cp,elementtype y){if((1)){printf(“溢出\n”);return NULL;}else{y=cp->element[cp->front];cp->front=(2);return y;}}

写出下列程序段的输出结果:栈结构定义为typedef struct astack *Stack;typedef struct astack{int top; //栈顶位置int maxtop; //栈顶位置的最大值StackItem *data; //栈元素数组指针}Astack;基本函数包括(1)初始化栈 StackInit (Stack S)(2)判断栈是否为空 int StackEmpty(Stack S)(3)判断栈是否满 int StackFull(Stack S)(4)入栈 Push(StackItem x,Stack S)(5)出栈 StackItem Pop(Stack S)(6)获取栈顶元素内容 StackItem StackTop(Stack S)void main( ) {Stack S;char x, y;x = 'c';y = 'k';Push ( x );Push ( 'a' );Push ( y );x=Pop ( S );Push ( 't' );Push ( x );x=Pop ( S );Push ( 's' );while (!StackEmpty ( ) ){y= Pop (S );printf(“%c”, y);}printf(“%c\n”, x);}输出结果:______________________

答案查题题库