数据结构实验之栈与队列三：后缀式求值



Problem Description

对于一个基于二元运算符的后缀表示式（基本操作数都是一位正整数），求其代表的算术表达式的值。

Input

输入一个算术表达式的后缀式字符串，以‘#’作为结束标志。

Output

求该后缀式所对应的算术表达式的值，并输出之。

Example Input

59*684/-3*+#

Example Output

57

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define stackmax 10000
#define stacknum 10000
typedef int element;
typedef struct
{
    element *base;
    element *top;
    int stacksize;
}sqstack;
int initstack(sqstack &S)
{
    S.base = (element *)malloc(stackmax *sizeof(element));
    if(!S.base) exit(0);
    S.top = S.base;
    S.stacksize = stackmax;
    return 0;
}
void push(sqstack &S, int e)
{
    if(S.top - S.base > S.stacksize)
    {
        S.base = (element *)realloc(S.base,(S.stacksize + stacknum) *sizeof(element));
        S.top = S.base + S.stacksize;
        S.stacksize += stacknum;
    }
    *S.top = e;
    S.top++;
}
int pop(sqstack &S,int e)
{
    if(S.base != S.top)
    {
        e = *(S.top-1);
        S.top--;
    }
    else
    {
        return 0;
    }
    return e;
}
int main()
{
    sqstack S;
    initstack(S);
    int e1, e2;
    char c;
    while(~scanf("%c", &c))
    {
        if(c == '#')
        {
            break;
        }
        if(c >= '0' && c <= '9')
        {
            push(S, c - '0');
        }
        else
        {
            if(c == '+')
            {
                e1 = pop(S, e1);
                e2 = pop(S, e2);
                push(S, e1+e2);
            }
            else if(c == '-')
            {
                e1 = pop(S, e1);
                e2 = pop(S, e2);
                push(S, e2-e1);
            }
            else if(c == '*')
            {
                e1 = pop(S, e1);
                e2 = pop(S, e2);
                push(S, e2*e1);
            }
            else if(c == '/')
            {
                e1 = pop(S, e1);
                e2 = pop(S, e2);
                push(S, e2/e1);
            }
        }
    }
    printf("%d\n", *(S.top-1));
    return 0;
}