int Pop(sqstack *s,int e) { if(s->top==s->base) return 0; e=*--s->top; return e;

int Pop(sqstack *s,int e) { if(s->top==s->base) return 0; e=*--s->top; return e; 一道数据结构题：#include #include #define STACK_INIT_SIZE 100#define STACKINCREMENT 10#define OK 1#define ERROR 0typedef struct{int *base;int *top;int stacksize;}SqStack;int InitStack(SqStack *S){S->base=(int *)malloc(STACK_INIT_SIZE*sizeof(i s.top = s.base + s.stacksize;之前是指针,到最后是怎么完成相加的?typedef struct sqstack{ int *base; int *top; int stacksize;}sqstack;sqstack s;之前是指针,到最后是怎么完成相加的? 进制转换#includestdio.h#define Max 50typedef struct{ int elem[Max]; int top;}Sqstack;void initstack(Sqstack*p){ (*p).top=-1;}int empty(Sqstack s){ if(s.top==-1) return 1; else ruturn 0;}void push(Sqstack*p,int x){ 如栈满,输入错误信息 其中的if(S.top-S.base >S.stacksize)是什么意思啊?int Push(SqStack &S,SElemType e) //Push() subfunction { //Push element to SqStack if(S.top-S.base >S.stacksize) //Stack == full?{ S.base=(SElemType *)realloc(S.base,(S.stacksize+STACKINCREMENT* 数据结构中sqstack（&s）和sqstack（s）有什么区别 C语言里if((i=!i)!void f31 (stack *S){Queue Q;Stack T;int i=0;InitQueue(&Q);InitStack(&T);While(!StackEmpty(S))if((i=!i)!=0) Push(&T,Pop(S));else EnQueue(&Q,Pop(S));.其中i指什么?书上说是S元素出栈,奇数入栈T,偶数入队Q,谢谢你 if((i=!i)!void f31 (stack *S){Queue Q;Stack T;int i=0;InitQueue(&Q);InitStack(&T);While(!StackEmpty(S))if((i=!i)!=0) Push(&T,Pop(S));else EnQueue(&Q,Pop(S));.书上说是S元素出栈,奇数入栈T,偶数入队Q, 一般算术表达式转化为逆波兰式#include#include#define SM 100typedef struct {char data[SM];int top;}Sqstack;void main(){Sqstack *sq;sq->top=0;int i=0,t=0;char ch,str[100],exp[100];printf(输入表达式（以#结尾）：);do {scanf(%c Pop_SqStack(&S,&k); /*回溯寻求下一个解*/ k怎么变化?#include#include #define MAXSIZE 100 /*最多元素数*/ typedef int ElemType; typedef struct{ ElemType data[MAXSIZE]; /*栈空间*/ int top; /*栈顶指针*/ }SqStack; void Init_SqStack( 括号匹配问题,出错在哪!#include#include#include#define OK 1#define ERROR -1#define OVERFLOW -2typedef struct{char *base;char *top;char stacksize;}SqStack;int InitStack(SqStack &s){s.base=(char *)malloc(100*sizeof(char));if(!s.base)exit(OVERF 出栈函数如下：Status Pop(SqStack &S,SElemType &e){if (S.top==S.base)return ERROR;e=*--S.top;return OK;}为什么是e=*--S.top; 而不是e=*S.top; e=*--S.top不就是取顶元素的下面那个元素么?有点混 int f() { static int i=0; int s=1; s+=i; i++; return s; }main() { int i,a=0; for(i=0;i f(int b[ ],int m,int n) {int i,s=0; for (i=m;i m=(int)(s[i]-'a')是什么意思 int i=1,s=1; for(;s+i void fun(char *s[],int n) { char *t;int i,j for(i=0;i 求下列程序算法的功能,希望尽量多给代码注释一下,多给点解释谢了bool function(char* rs){if(!rs || !rs[0])return false;int nLen = strlen(rs);//获得rs的长度SqStack S;InitStack(S, nLen);bool ret = true;for(char *p = str; p