(define producer #f)
(define (produce)
(let loop ((i 0))
(call/cc (lambda (k)
(define item (string-append "item-" (number->string i)))
(display "Producing ") (display item) (newline)
(set! producer k)
(consume item)))
(loop (1+ i))))#include <stdio.h>
#include <math.h>
long fun(long s)
{
long n = 0, t1;
long t = 0;
while (s)
{
s /= 10;#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <fcntl.h>
#include <pthread.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <liburing.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <fcntl.h>
#include <errno.h>
#include <pthread.h>
#include <sys/socket.h>
#include <sys/epoll.h>
#include <netinet/in.h>#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <fcntl.h>
#include <errno.h>
#include <pthread.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <pthread.h>
#include <sys/socket.h>
#include <netinet/in.h>
#define PORT 8080
#define BUFFER_SIZE 1024// 编写一个函数fun,然后设计主函数调用函数fun。函数fun的功能是:求出两个非零正整数的最大公约数(考虑递归和非递归两种方法,任选一种实现),
// 并作为函数值返回。再设计一个函数,函数的功能是计算两个非零正整数的最小公倍数。
#include <stdio.h>
int min(int, int, int);
int fun(int, int, int);
int main(void)
{
int a, b, max;#include <stdio.h>
#include <math.h>
int fun(int);
int main(void)
{
int a, n, result;
scanf("%d%d", &a, &n);
result = a * fun(n);
for (int i = n - 1; i > 0; i--)#include <stdio.h>
#include <stdlib.h>
int generateNumber(const char, const int);
int main(void)
{
char a;
int n;
scanf("%c%d", &a, &n);#include <cmath>
#include <iostream>
#include <list>
class Term {
public:
int coef;
int exp;
};