编写包含栈的程序，完成手机计算机器中的简单表达式求值，完成+、-、*、/、% 五种混合运算和以下要求的C语言代码 1.无括号、整数 2.有括号、整数 3.有括号、多位整数

以下是实现上述要求的C语言代码：

#include <stdio.h>
#include <stdlib.h>

#define MAX_SIZE 100 

typedef struct {
    int top;
    int data[MAX_SIZE];
} Stack;

void init(Stack *s) {
    s->top = -1;
}

int is_empty(Stack *s) {
    return s->top == -1;
}

int is_full(Stack *s) {
    return s->top == MAX_SIZE - 1;
}

void push(Stack *s, int x) {
    if (is_full(s)) {
        printf("Stack overflow\n");
        exit(1);
    }
    s->data[++s->top] = x;
}

int pop(Stack *s) {
    if (is_empty(s)) {
        printf("Stack underflow\n");
        exit(1);
    }
    return s->data[s->top--];
}

int peek(Stack *s) {
    if (is_empty(s)) {
        printf("Stack underflow\n");
        exit(1);
    }
    return s->data[s->top];
}

int is_operator(char c) {
    return c == '+' || c == '-' || c == '*' || c == '/' || c == '%';
}

int precedence(char op) {
    if (op == '+' || op == '-') {
        return 1;
    } else if (op == '*' || op == '/' || op == '%') {
        return 2;
    } else {
        return 0;
    }
}

int evaluate_expression(char *exp) {
    Stack operand_stack;
    Stack operator_stack;
    init(&operand_stack);
    init(&operator_stack);
    int i = 0;
    while (exp[i] != '\0') {
        if (exp[i] == '(') {
            push(&operator_stack, exp[i]);
        } else if (exp[i] == ')') {
            while (!is_empty(&operator_stack) && peek(&operator_stack) != '(') {
                char op = pop(&operator_stack);
                int b = pop(&operand_stack);
                int a = pop(&operand_stack);
                int result;
                switch (op) {
                    case '+':
                        result = a + b;
                        break;
                    case '-':
                        result = a - b;
                        break;
                    case '*':
                        result = a * b;
                        break;
                    case '/':
                        result = a / b;
                        break;
                    case '%':
                        result = a % b;
                        break;
                }
                push(&operand_stack, result);
            }
            pop(&operator_stack);
        } else if (is_operator(exp[i])) {
            while (!is_empty(&operator_stack) && peek(&operator_stack) != '(' && precedence(exp[i]) <= precedence(peek(&operator_stack))) {
                char op = pop(&operator_stack);
                int b = pop(&operand_stack);
                int a = pop(&operand_stack);
                int result;
                switch (op) {
                    case '+':
                        result = a + b;
                        break;
                    case '-':
                        result = a - b;
                        break;
                    case '*':
                        result = a * b;
                        break;
                    case '/':
                        result = a / b;
                        break;
                    case '%':
                        result = a % b;
                        break;
                }
                push(&operand_stack, result);
            }
            push(&operator_stack, exp[i]);
        } else {
            int num = 0;
            while (exp[i] >= '0' && exp[i] <= '9') {
                num = num * 10 + exp[i] - '0';
                i++;
            }
            push(&operand_stack, num);
            continue;
        }
        i++;
    }
    while (!is_empty(&operator_stack)) {
        char op = pop(&operator_stack);
        int b = pop(&operand_stack);
        int a = pop(&operand_stack);
        int result;
        switch (op) {
            case '+':
                result = a + b;
                break;
            case '-':
                result = a - b;
                break;
            case '*':
                result = a * b;
                break;
            case '/':
                result = a / b;
                break;
            case '%':
                result = a % b;
                break;
        }
        push(&operand_stack, result);
    }
    return pop(&operand_stack);
}

int main() {
    char exp[MAX_SIZE];
    printf("Enter expression: ");
    scanf("%s", exp);
    int result = evaluate_expression(exp);
    printf("Result: %d\n", result);
    return 0;
}

注意：该程序只支持整数运算，不支持浮点数运算。如果需要支持浮点数运算，请修改相关代码。