利用线程方案实现实现哲学家问题.设置五个信号量代表五把叉子，初使值均为1，表示5把叉子均可以取，设置五个线程代表5个哲学家，其值分别为0～4，规定奇数号的哲学家先拿起他左边的叉子,然后再去拿他右边的叉子 而偶数号的哲学家则相反.按此规定,将是1,2号哲学家竞争1号叉子,3,4号哲学家竞争3号叉子.即五个哲学家都竞争奇数号叉子,获得后,再去竞争偶数号叉子,最后总会有一个哲学家能获得两支叉子而进餐。而申请不到的哲学家进入阻塞等待队列，根FIFO原则，则先申请的哲学家会较先可以吃饭，因此不会出现饿死的哲学家。-use threaded program to achieve realization philosophers problem. Installed capacity on behalf of five signal five fork, in early values are one, five fork can take, and set up five representatives of five threads philosopher, the value of 0 to 4, provisions of the odd philosopher he first took up the left fork. then plain his right fork, and even philosophers, on the contrary. accordingly, will be one, on the 2nd philosopher competition on the 1st fork, 3, philosopher competition on the 4th on the 3rd fork. that is five philosophers competition odd, fork, access, even longer competitive, fork, they will eventually have access to a philosopher two fork and eat. But philosophers are not eligible to enter the queue waiting for obstruction, root FIFO principle, I will apply the first philosop