编制具有如下原型的函数prime，用来判断整数n是否为素数：bool prime(int n) 而后编制主函数，任意输入一个大于4的偶数d，找出满足d=d1+d2的所有数对，其中要求d1与d2均为素数（通过调用prime来判断素数）。如偶数18可以分解为11+7以及13+5；而偶数80可以分解为：43+37、61+19、67+13、73+7。

　　提示：i与d-i的和恰为偶数d，而且只有当i与d-i均为奇数时才有可能成为所求的“数对”。

-Prepared with the following function prototype prime, used to determine whether an integer n prime numbers: bool prime (int n) before the preparation of the main function, indiscriminate importation of a greater than 4, even d, to find out to satisfy d = d1+ D2 all the number of , which calls for d1 and d2 are prime numbers (by calling the prime to determine prime numbers). If even 18 can be decomposed into 11+ 7 and 13+ 5　and even 80 can be decomposed into: 43+ 37,612 B! 19,672 B! 13,732 B! 7. Tip: i with di and exactly even d, and only when i and di are odd when they may become the order of　a few of.