设某银行有A、B两个业务窗口，且处理业务的速度不一样，其中A窗口处理速度是B窗口的2倍+——+即当A窗口每处理完2个顾客时，B窗口处理完1个顾客。给定到达银行的顾客序列请按业务完成的顺序输出顾客序列，用c语言编译程序

这是一个典型的银行排队问题，可以使用队列来解决。我们可以将顾客按照到达时间依次加入队列中，然后根据A、B窗口的处理速度，依次从队列中取出顾客进行处理。具体实现可以参考以下代码：

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

#define MAX_SIZE 100

typedef struct {
    int id; // 顾客编号
    int time; // 顾客需要处理的时间
} Customer;

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

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

int isEmpty(Queue *q) {
    return q->front == q->rear;
}

int isFull(Queue *q) {
    return (q->rear + 1) % MAX_SIZE == q->front;
}

void enqueue(Queue *q, Customer c) {
    if (isFull(q)) {
        printf("Queue is full!\n");
        exit(1);
    }
    q->data[q->rear] = c;
    q->rear = (q->rear + 1) % MAX_SIZE;
}

Customer dequeue(Queue *q) {
    if (isEmpty(q)) {
        printf("Queue is empty!\n");
        exit(1);
    }
    Customer c = q->data[q->front];
    q->front = (q->front + 1) % MAX_SIZE;
    return c;
}

int main() {
    Queue qA, qB;
    initQueue(&qA);
    initQueue(&qB);

    int n, id, time;
    printf("Enter the number of customers: ");
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        printf("Enter the id and time of customer %d: ", i + 1);
        scanf("%d%d", &id, &time);
        Customer c = {id, time};
        enqueue(&qA, c);
    }

    int curTime = 0;
    while (!isEmpty(&qA) || !isEmpty(&qB)) {
        if (!isEmpty(&qA)) {
            Customer c = dequeue(&qA);
            printf("Processing customer %d at A window\n", c.id);
            curTime += c.time;
            if (isEmpty(&qA) || curTime % 2 == 0) {
                if (!isEmpty(&qB)) {
                    Customer c = dequeue(&qB);
                    printf("Processing customer %d at B window\n", c.id);
                    curTime += c.time;
                }
            } else {
                enqueue(&qB, c);
            }
        } else {
            Customer c = dequeue(&qB);
            printf("Processing customer %d at B window\n", c.id);
            curTime += c.time;
        }
    }

    return 0;
}