用C语言实现：【问题描述】 回文判断：假设称正读和反读都相同的字符序列为“回文”，例如‘abba’和‘abcba’是回文，‘abcde’和‘ababab’则不是。试写一个算法判别读入的一个以‘@’结束符的字符序列是否是“回文”。 提示：利用栈和队列知识。 Copying judgment: Assuming that the character sequence known as the same reading and anti-reading is "replying", such as 'abba' and 'abcba' are repayment, 'Abcde' and 'Ababab' are not. Try to write a character sequence of an algorithm to read in the '@' ending character. Tip: Use stacks and queue knowledge. 【输入形式】 以@结尾的字符串 【输出形式】 判断结果 【样例输入】 abcdcba@ 【样例输出】 此字符串为回文

以下是C语言实现的代码：

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

#define MAX_SIZE 100

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

typedef struct {
    char data[MAX_SIZE];
    int front;
    int rear;
} Queue;

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

void Push(Stack *s, char ch) {
    if (s->top == MAX_SIZE - 1) {
        printf("Stack overflow!");
        exit(1);
    }
    s->data[++s->top] = ch;
}

char Pop(Stack *s) {
    if (s->top == -1) {
        printf("Stack underflow!");
        exit(1);
    }
    return s->data[s->top--];
}

void InitQueue(Queue *q) {
    q->front = q->rear = 0;
}

void EnQueue(Queue *q, char ch) {
    if (q->rear == MAX_SIZE - 1) {
        printf("Queue overflow!");
        exit(1);
    }
    q->data[q->rear++] = ch;
}

char DeQueue(Queue *q) {
    if (q->front == q->rear) {
        printf("Queue underflow!");
        exit(1);
    }
    return q->data[q->front++];
}

int main() {
    char str[MAX_SIZE];
    Stack s;
    Queue q;
    int i, len, is_palindrome = 1;

    printf("Please input a string with @ as the ending character:\n");
    scanf("%s", str);

    len = strlen(str);

    InitStack(&s);
    InitQueue(&q);

    for (i = 0; i < len - 1; i++) {
        Push(&s, str[i]);
        EnQueue(&q, str[i]);
    }

    while (s.top != -1 && q.front != q.rear) {
        if (Pop(&s) != DeQueue(&q)) {
            is_palindrome = 0;
            break;
        }
    }

    if (is_palindrome) {
        printf("This string is a palindrome.\n");
    } else {
        printf("This string is not a palindrome.\n");
    }

    return 0;
}

该算法使用栈和队列来实现，先将输入字符串中除了结尾的'@'之外的字符依次入栈和入队，然后依次弹出栈顶和队首的字符进行比较，如果出现不相等的情况，则说明该字符串不是回文。