任意一个大于2的偶数，都可以分解为两个质数之和。编写一个程序

#include<iostream>
#include<cmath>
using namespace std;
bool isPrime( int num )
{
    //两个较小数另外处理
    if(num ==2|| num==3 )
        return 1 ;
    //不在6的倍数两侧的一定不是质数
    if(num %6!= 1&&num %6!= 5)
        return 0 ;
    int tmp =sqrt( num);
    //在6的倍数两侧的也可能不是质数
    for(int i= 5; i <=tmp; i+=6 )
        if(num %i== 0||num %(i+ 2)==0 )
            return 0 ;
    //排除所有，剩余的是质数
    return 1 ;
}
int main()
{
    int x;
    cin>>x;
    int i=2,j=x-i;
    if(x%2==0){
        while(i<=j){
            if(isPrime(i)&&isPrime(j)){
                cout<<x<<"="<<i<<"+"<<j<<endl;
                return 0;
            }
            i++;j--;
        }
        cout<<"Can't do it."<<endl;
    }else{
        cout<<"The number is not even."<<endl;
    }
    return 1;
}

import random
def zhi_number(k):#判断是否是质数
    flag=True
    beichushu=2
    while beichushu<=k/2:
        if k%beichushu==0:
            flag=False
            break
        else:
            beichushu+=1
    return flag
def qiuzhishu(n):#求某个数的质数集合列表
    zhishu_list=list()
    chushu=2
    while chushu<n:
        if zhi_number(chushu)==True:
            zhishu_list.append(chushu)
        chushu+=1
    return zhishu_list
def yanzheng():
    name=random.randint(1,2**32)
    while name%2!=0:
        name=random.randint(1,2**32)
    # 随机生成一个偶数，范围很大，2的32次方
    zhishuset=qiuzhishu(name)
    falg=False
    for name1 in zhishuset:#在质数集合里找第一个加数
        for name2 in zhishuset:#找第二个加数
            if name==name1+name2:#判断想加是否等于随机生成的偶数
                falg=True
                break
    return falg
print(yanzheng())

#include<math.h>
#include<stdio.h>

int isprime(int x)
{//判断一个数是否是质数
    int start=2;
    while(start*start<=x)
    {//只需要试探到根号x就已经足够
        if(x%start==0)
            return 0;
        else
            start++;
    }
    return 1;
}

int main()
{
    int num,explore = 2,sub_explore;//explore从最小的质数开始试探
    printf("input an even number:");
    scanf("%d",&num);
    while(num%2)//如果输入的不是偶数就请求一个偶数
    {
        printf("erro!need an even number:\n");
        scanf("%d",&num);
    }
    sub_explore = num-explore;
    while(isprime(explore) && isprime(sub_explore))
        //用while循环寻找两个质数之和
    {
        explore++;
        sub_explore--;
        if(sub_explore>explore)//定理被推翻!!!
            printf("niubi!!!");
    }
    printf("%d can be broken down into %d and %d: ",num,explore,sub_explore);
    return 1;
}

def primality(n):
    if n <= 102:
        for a in range(2, n):
            if pow(a, n - 1, n) != 1:
                return False
    else:
        import random
        for i in range(100):
            a = random.randint(2, n - 1)
            if pow(a, n - 1, n) != 1:
                return False
    return True

while True:
    try:
        A = int(input('Please input an even number:').strip())
        if A=='':
            break
        elif A%2!=0:
            print('Not even, please retry!')
            continue
        if A==4:
            print('4 = 2 + 2')
            continue
        halfA,flag = int(A/2),False
        # just need to check odd numbers, >=3
        for S in range(3,halfA+1,2):
            B = A - S
            if primality(S) and primality(B):
                flag = True
                print('%d = %d + %d' %(A,S,B))
        if not flag:
            print('%d cannot be divided!' %A)
    except:
        break

	
	

		//任意大于2的偶数，都可以分解成两个质数之和
	

	

		#include<stdio.h>
	

	

		int Su(int n){
	

	

		int answer=1;
	

	

		for(int temp=2;temp*temp<n;temp++){
	

	

		if(n%temp==0)
	

	

		answer=0;
	

	

		}
	

	

		return answer;
	

	

		}
	

	

		int main(void){
	

	

		int a;
	

	

		printf("Please Input a Number:\n");
	

	

		scanf("%d",&a);
	

	

		while(a%2!=0||a<=2){
	

	

		printf("Wrong Input! Repeat:");
	

	

		scanf("%d",&a);
	

	

		}
	

	

		
	

	

		for(int temp=1;temp<a;temp++){
	

	

		if(Su(temp)&&Su(a-temp)){
	

	

		printf("Answer:%d %d\n",temp,a-temp);
	

	

		break;
	

	

		}
	

	

		}
	

	

		return 0;
	

	

		}