Conversion expression from infix to suffix

Conversion expression from infix to suffix (15 points)
Grade: 15/discount: 0.8
Problem description:
An infix expression is a mathematical expression that we usually write. A suffix expression is also called an inverse polish expression. When compiling a program to check the syntax of the expressions in the program we write, it is often possible to do so through a reverse polish expression. The program we want to design and implement is to convert the arithmetic expression represented by infix into a suffix, for example,
Replace the infix expression
(A 1 (B * C 10 d) * E)/(f 10g)
Convert to suffix
ABC * d Ten E *-FG ten/
Note: to simplify programming, assuming that the variable name is a positive real number and the operator is only +,-, *, And/, You can process the delimiter and assume that the input arithmetic expression is correct.
Requirement: Use the stack data structure. The input infix expression ends #.
Input:
Integer n. There are n infix expressions below
N expressions composed of positive numbers and operators
Output
N suffix expressions. The operands and operators are separated by spaces.

 

#include <stdio.h>#include <stdlib.h>struct shu{    int front;    int rear;}OPND;struct others {    char * base;    char * top;    int size;}OPTR;char bijiao(char x, char y) {    if (x == '+') {        if (y == '+' || y == '-' || y == ')' || y == '#')            return('>');        else            return('<');    }    if (x == '-') {        if (y == '+' || y == '-' || y == ')' || y == '#')            return('>');        else            return('<');    }    if (x == '*') {        if (y == '(')            return('<');        else            return('>');    }    if (x == '/') {        if (y == '(')            return('<');        else            return('>');    }    if (x == '(') {        if (y == ')')            return('=');        if (y == '+' || y == '-' || y == '*' || y == '/' || y == '%' || y == '(')            return('<');    }    if (x == ')') {        if (y == '+' || y == '-' || y == '*' || y == '/' || y == '%' || y == ')' || y == '#')            return('>');    }    if (x == '#') {        if (y == '+' || y == '-' || y == '*' || y == '/' || y == '%' || y == '(')            return('<');        if (y == '#')            return('=');    }}void main() {    int n, i, j, k, s[50], flag;    char sopnd[50][50], str[50],t;    scanf("%d", &n);    getchar();    for (i = 0;i < n;i++) {        flag = 0;        OPND.front = 0;        OPND.rear = 0;        OPTR.size = 50;        OPTR.base = (char *)malloc(OPTR.size * sizeof(char));        OPTR.top = OPTR.base;        *OPTR.top++ = '#';        for (j = 0;j < 50;j++)            s[j] = 0;        gets(str);        for (j = 0;str[j] != '#' || *(OPTR.top - 1) != '#';j++) {            if (str[j] >= '0' && str[j] <= '9') {                for (k = 0;(str[j] <= '9' && str[j] >= '0') || str[j] == '.';) {                    sopnd[OPND.rear][k] = str[j];                    j++;                    k++;                }                sopnd[OPND.rear][k] = '\0';                OPND.rear++;                j--;            }              else {              t= bijiao(*(OPTR.top - 1), str[j]); {                if(t=='<') {                    *OPTR.top++ = str[j];                                    }                if(t=='='){                    OPTR.top--;                                    }                if(t=='>') {                    for (;OPND.front < OPND.rear;OPND.front++) {                        if (flag == 1)                            printf(" %s", sopnd[OPND.front]);                        else                            printf("%s", sopnd[OPND.front]);                        flag = 1;                    }                    printf(" %c", *--OPTR.top);                    j--;                                    }                }            }        }        printf("\n");    }}