如：

通过N个线程顺序循环打印从0至100，如给定N=3则输出:thread0: 0thread1: 1thread2: 2thread0: 3thread1: 4.....

解：N个线程对应N个条件变量，依次激活下一个线程

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         constexpr int THREADS = 3;constexpr int MAX = 100;std::mutex m;std::condition_variable cv[THREADS];bool flags[THREADS];int n = 0;void f(int i){    const int next = (i + 1) % THREADS;    for (;;)    {        std::unique_lock
         
           lock(m); cv[i].wait(lock, [=](){return flags[i];}); if (n <= MAX) std::cout << "thread " << i << ": " << n++ << std::endl; flags[i] = false; flags[next] = true; lock.unlock(); cv[next].notify_one(); if (n > MAX) break; }}int main(){ flags[0] = true; std::vector
          
            vec; for (int i = 0; i < THREADS; ++i) vec.emplace_back(f, i); for (auto &t: vec) t.join(); return 0;}