题目链接：https://nanti.jisuanke.com/t/25985
题目：
Description:
Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states:
Every even integer greater than 2 can be expressed as the sum of two primes.
The actual verification of the Goldbach conjecture shows that even numbers below at least 1e14 can be expressed as a sum of two prime numbers.
Many times, there are more than one way to represent even numbers as two prime numbers.
For example, 18=5+13=7+11, 64=3+61=5+59=11+53=17+47=23+41, etc.
Now this problem is asking you to divide a postive even integer n (2<n<2^63) into two prime numbers.
Although a certain scope of the problem has not been strictly proved the correctness of Goldbach's conjecture, we still hope that you can solve it.
If you find that an even number of Goldbach conjectures are not true, then this question will be wrong, but we would like to congratulate you on solving this math problem that has plagued humanity for hundreds of years.
Input:
The first line of input is a T means the number of the cases.
Next T lines, each line is a postive even integer n (2<n<2^63).
Output:
The output is also T lines, each line is two number we asked for.
T is about 100.
本题答案不唯一，符合要求的答案均正确
样例输入
1 8
样例输出
3 5
题意：哥德巴赫猜想：任意一个大于2的偶数即可表示为两个素数的和。运用Miller_Rabin判别法从n/2向两边遍历（由素数的分布可知这样时间复杂度更优）。
代码实现如下：
1 #include <cstdio> 2 #include <ctime> 3 #include <cstdlib> 4 5 typedef long long ll; 6 int t; 7 long long n; 8 9 ll multi(ll a, ll b, ll mod) { 10 ll ret = 0; 11 while(b) { 12 if(b & 1) 13 ret = ret + a; 14 if(ret >= mod) 15 ret -= mod; 16 17 a = a + a; 18 if(a >= mod) 19 a -= mod; 20 b >>= 1; 21 } 22 return ret; 23 } 24 ll quick_pow(ll a, ll b, ll mod) { 25 ll ret = 1; 26 while(b) { 27 if(b & 1) 28 ret = multi(ret, a, mod); 29 a = multi(a, a, mod); 30 b >>= 1; 31 } 32 return ret; 33 } 34 bool Miller_Rabin(ll n) { 35 ll u = n - 1, pre, x; 36 int i, j, k = 0; 37 if(n == 2 || n == 3 || n == 5 || n == 7 || n == 11) 38 return true; 39 if(n == 1 || (!(n % 2)) || (!(n % 3)) || (!(n % 5)) || (!(n % 7)) || (!(n % 11))) 40 return false; 41 for(; !(u & 1); k++, u >>= 1); 42 srand(time(NULL)); 43 for(i = 0; i < 5; i++) { 44 x = rand() % (n - 2) + 2; 45 x = quick_pow(x, u, n); 46 pre = x; 47 for(j = 0; j < k; j++) { 48 x = multi(x, x, n); 49 if(x == 1 && pre != 1 && pre != (n - 1)) 50 return false; 51 pre = x; 52 } 53 if(x != 1) 54 return false; 55 } 56 return true; 57 } 58 59 int main() { 60 scanf("%d", &t); 61 while(t--) { 62 scanf("%lld", &n); 63 long long k = n /2; 64 if(Miller_Rabin(k)) { 65 printf("%lld %lld ", k, k); 66 } else { 67 long long l = k - 1, r = k + 1; 68 while(!Miller_Rabin(l) || !Miller_Rabin(r)) { 69 l--; 70 r++; 71 } 72 printf("%lld %lld ", l, r); 73 } 74 } 75 return 0; 76 }