时间片轮转调度算法与银行家算法结合，设计一种时间片轮转调度算法从进程就绪队列中选择一个进程，该进程发出资源请求并启动银行家算法，打印文本为中文。给出c语言实现代码

好的，以下是我设计的时间片轮转调度算法与银行家算法结合的代码实现：

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

#define MAX_RESOURCES 5 // 系统资源数
#define MAX_PROCESSES 3 // 系统进程数

int available_resources = MAX_RESOURCES;
int allocation[MAX_PROCESSES] = {0, 0, 0}; // 各进程已分配资源数
int need[MAX_PROCESSES] = {2, 3, 1}; // 各进程还需要资源数

typedef struct Process {
    int pid; // 进程ID
    int remaining_time; // 剩余时间片
    int requested_resources; // 请求资源数
    int allocated_resources; // 已分配资源数
    int need_resources; // 还需要资源数
} Process;

Process* create_process(int pid, int remaining_time, int requested_resources) {
    Process* process = (Process*) malloc(sizeof(Process));
    process->pid = pid;
    process->remaining_time = remaining_time;
    process->requested_resources = requested_resources;
    process->allocated_resources = 0;
    process->need_resources = requested_resources;
    return process;
}

void run_process(Process* process) {
    printf("正在运行进程 %d...\n", process->pid);
    process->remaining_time--;
    if (process->remaining_time == 0) {
        printf("进程 %d 运行完毕！\n", process->pid);
        available_resources += process->allocated_resources;
        need[process->pid] = 0;
    }
}

int is_safe_state(int pid, int requested_resources) {
    if (requested_resources > available_resources) {
        return 0; // 若请求资源数大于系统资源数，则不安全
    }
    int new_allocation[MAX_PROCESSES];
    int new_need[MAX_PROCESSES];
    for (int i = 0; i < MAX_PROCESSES; i++) {
        new_allocation[i] = allocation[i];
        new_need[i] = need[i];
    }
    new_allocation[pid] += requested_resources;
    new_need[pid] -= requested_resources;
    int work[MAX_RESOURCES];
    for (int i = 0; i < MAX_RESOURCES; i++) {
        work[i] = available_resources;
    }
    int finish[MAX_PROCESSES] = {0, 0, 0};
    int count = 0;
    while (count < MAX_PROCESSES) {
        int found = 0;
        for (int i = 0; i < MAX_PROCESSES; i++) {
            if (!finish[i] && new_need[i] <= work[0] && new_need[i + 1] <= work[1] && new_need[i + 2] <= work[2]) {
                work[0] += new_allocation[i];
                work[1] += new_allocation[i + 1];
                work[2] += new_allocation[i + 2];
                finish[i] = 1;
                count++;
                found = 1;
            }
        }
        if (!found) {
            break;
        }
    }
    return count == MAX_PROCESSES;
}

int main() {
    Process* processes[MAX_PROCESSES];
    processes[0] = create_process(0, 3, 2);
    processes[1] = create_process(1, 2, 3);
    processes[2] = create_process(2, 4, 1);
    printf("初始状态：\n");
    printf("可用资源数：%d\n", available_resources);
    printf("进程\t已分配\t还需求\n");
    for (int i = 0; i < MAX_PROCESSES; i++) {
        printf("%d\t%d\t%d\n", i, allocation[i], need[i]);
    }
    printf("\n");
    while (1) {
        for (int i = 0; i < MAX_PROCESSES; i++) {
            if (need[i] == 0) {
                continue;
            }
            if (processes[i]->remaining_time == 0) {
                processes[i]->remaining_time = 3;
            }
            if (processes[i]->remaining_time > 0) {
                printf("进程 %d 正在运行...\n", processes[i]->pid);
                run_process(processes[i]);
                if (processes[i]->remaining_time == 0) {
                    printf("释放进程 %d 占用的资源...\n", processes[i]->pid);
                    available_resources += processes[i]->allocated_resources;
                    allocation[processes[i]->pid] = 0;
                    need[processes[i]->pid] = 0;
                }
            } else {
                if (is_safe_state(i, processes[i]->requested_resources)) {
                    printf("进程 %d 在请求 %d 个资源...\n", processes[i]->pid, processes[i]->requested_resources);
                    allocation[processes[i]->pid] += processes[i]->requested_resources;
                    available_resources -= processes[i]->requested_resources;
                    processes[i]->allocated_resources += processes[i]->requested_resources;
                    processes[i]->need_resources -= processes[i]->requested_resources;
                    if (processes[i]->need_resources == 0) {
                        printf("进程 %d 已经获得了所有需要的资源！\n", processes[i]->pid);
                        need[processes[i]->pid] = 0;
                    }
                } else {
                    printf("进程 %d 请求资源失败，等待下一次调度...\n", processes[i]->pid);
                }
            }
        }
        int flag = 1;
        for (int i = 0; i < MAX_PROCESSES; i++) {
            if (need[i] > 0) {
                flag = 0;
                break;
            }
        }
        if (flag) {
            printf("\n所有进程都已运行完毕！\n");
            break;
        }
    }
    return 0;
}

以上代码实现了一个包含三个进程的系统，每个进程需要一定数量的资源才能运行。时间片大小为3，当一个进程的时间片用完后，它将释放已分配的资源，并请求更多资源，直到它获得所有需要的资源为止。在请求资源时，将启动银行家算法来判断是否安全。如果请求资源失败，则该进程将等待下一次调度。当所有进程都运行完毕时，程序结束。

希望这个实现能够满足您的需求！