刚看到ACM的一道赛题，很简单：
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
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
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
整体来说就是去寻找素数，所以，这个ACM赛题我觉得就是在考察寻找素数的算法。而这个算法简单的来说，就是挨个的进行尝试，但是很显然，这是很慢的。即使你只是循环到sqrt（number）;
在网上看到了一种新的方式，对于我来说是一种新的方式，
就是说，对于一些比5大的素数，她们都是在6的倍数的上下分布的，比如5.7 11 13 17 19
不一而是。根据这个想法，很简单的就能写出一种新的算法：下面就是我这道题的C++代码实现
#include <iomanip>
#include <math.h>
#include <iostream>
using namespace std;
bool judge(int num){
if(num ==2|| num==3 ){//这是对比较小的数进行排除
return 1 ;
}
if(num %6!= 1&&num %6!= 5){ //然后进行判断是不是在6的两侧，在六的两测是素数的必要不充分条件
return 0 ;
}
int index =sqrt( num);
for(int i= 5;i <=index; i+=6 ){ //所以说在6的倍数两侧的也可能不是质数，这样就简单多了。循环的次数少多了
return ((num %i== 0||num %(i+ 2)==0)==1?0:1);
}
}
void FindAndOut(int number){
if(judge(2)&&judge(number-2)){
cout<<setiosflags(ios::left)<<setw(4)<<2<<(number-2)<<endl;
}
for(int i=3;i<(number/2);i+=2){
if(judge(i)&&judge(number-i)){
cout<<setiosflags(ios::left)<<setw(4)<<i<<(number-i)<<endl;
break;
}
}
}
using namespace std;
int main(){
int n;
int *num;
cin>>n;
num = new int[n];
for(int i=0;i<n;i++){
cin>>num[i];
}
for(int i=0;i<n;i++){
FindAndOut(num[i]);
}
delete []num;
return 0;
}