产生式为L→En E→E1+T E→T T→T1F T→F F→E F→1 F→2 F→3 F→4 F→5利用LR分析法和语法制导翻译法实现简单算术表达式求值。例如：输入35+4+2n 输出29。生成完整c语言代码编写正确的代码能通过编译器验证

以下是完整的C语言代码实现：

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

#define MAX_STACK_SIZE 100
#define MAX_INPUT_SIZE 100

// 定义产生式
char productions[10][10] = {"L->En", "E->E1+T", "E->T", "T->T1*F", "T->F", "F->(E)", "F->1", "F->2", "F->3", "F->4", "F->5"};

// 定义符号栈结构体
typedef struct {
    char stack[MAX_STACK_SIZE];
    int top;
} Stack;

// 初始化符号栈
void initStack(Stack *s) {
    s->top = -1;
}

// 判断符号栈是否为空
int isEmpty(Stack *s) {
    return s->top == -1;
}

// 判断符号栈是否已满
int isFull(Stack *s) {
    return s->top == MAX_STACK_SIZE - 1;
}

// 入栈
void push(Stack *s, char c) {
    if (isFull(s)) {
        printf("Error: Stack is full.\n");
        exit(EXIT_FAILURE);
    }
    s->stack[++s->top] = c;
}

// 出栈
char pop(Stack *s) {
    if (isEmpty(s)) {
        printf("Error: Stack is empty.\n");
        exit(EXIT_FAILURE);
    }
    return s->stack[s->top--];
}

// 读取输入字符串
void readInput(char *input) {
    printf("请输入要求值的算术表达式：");
    scanf("%s", input);
}

// 获取下一个输入字符
char getNextChar(char *input, int *pos) {
    return input[(*pos)++];
}

// 将字符转为数字
int charToInt(char c) {
    return c - '0';
}

// 计算表达式的值
int calculate(Stack *s, char *input) {
    int pos = 0;
    char lookahead = getNextChar(input, &pos);
    push(s, '$');
    push(s, 'L');
    while (lookahead != '$' && !isEmpty(s)) {
        char top = s->stack[s->top];
        if (top == lookahead) {
            pop(s);
            lookahead = getNextChar(input, &pos);
        } else if (isupper(top)) {
            int i, j;
            for (i = 0; i < 11; i++) {
                if (top == productions[i][0]) {
                    for (j = strlen(productions[i]) - 1; j >= 3; j--) {
                        push(s, productions[i][j]);
                    }
                    break;
                }
            }
        } else if (isdigit(top)) {
            int num = charToInt(top);
            pop(s);
            while (!isEmpty(s) && s->stack[s->top] == 'T') {
                pop(s);
                char op = pop(s);
                int operand = charToInt(pop(s));
                if (op == '*') {
                    num *= operand;
                } else if (op == '/') {
                    num /= operand;
                }
            }
            push(s, num + '0');
        } else {
            printf("Error: Invalid input.\n");
            exit(EXIT_FAILURE);
        }
    }
    return charToInt(s->stack[s->top]);
}

int main() {
    char input[MAX_INPUT_SIZE];
    readInput(input);
    Stack s;
    initStack(&s);
    int result = calculate(&s, input);
    printf("计算结果为：%d\n", result);
    return 0;
}
``