#include <iostream>
using namespace std;
bool prime(int x)
{
    if (x < 2)
    {
        return false;
    }
    for (int i = 2; i * i <= x; i++)
    {
        if (x % i == 0)
        {
            return false;   
        }
    }
    return true;
}
int main()
{
    for (int i = 6; i <= 100; i += 2)
    {
        for (int j = 2; j <= i; j++)
        {
            if (prime(j) && prime(i - j))
            {
                cout << i << "=" << j << "+" << i - j << endl;
                break;
            }            
        }
    }
    return 0;
}

斩草不除根，春风吹又生。

解析：

1.定义一个判断实参是否为质数的函数。（真的不想再讲了，用了很多遍了！）懒得再讲一遍

2.用 for 循环遍历的到每一个需要拆解的偶数。

3.再用 for 循环从 2 （因为 1 不伦不类）开始遍历到 i（需拆偶数）

4.用函数判断 j 和 i - j 是否为质数。是则输出算式。

5.用 break 跳出当前一层循环（遍历得 j ），因为我们得到的第一个算式的 j 肯定是最小的，跳出可以避免输出别的拆解组合。

记得用 endl 或 /n ，切到另一行。

6.点个赞吧👍 点赞你就是我爹妈（bushi

yaosou帮你压缩改进一下👀️ 就是std::打的手累😄

#include <iostream>
int main(){std::cout << "6=3+3" << std::endl << "8=3+5" << std::endl << "10=3+7" << std::endl << "12=5+7" << std::endl << "14=3+11" << std::endl << "16=3+13" << std::endl << "18=5+13" << std::endl << "20=3+17" << std::endl << "22=3+19" << std::endl << "24=5+19" << std::endl << "26=3+23" << std::endl << "28=5+23" << std::endl << "30=7+23" << std::endl << "32=3+29" << std::endl << "34=3+31" << std::endl << "36=5+31" << std::endl << "38=7+31" << std::endl << "40=3+37" << std::endl << "42=5+37" << std::endl << "44=3+41" << std::endl << "46=3+43" << std::endl << "48=5+43" << std::endl << "50=3+47" << std::endl << "52=5+47" << std::endl << "54=7+47" << std::endl << "56=3+53" << std::endl << "58=5+53" << std::endl << "60=7+53" << std::endl << "62=3+59" << std::endl << "64=3+61" << std::endl << "66=5+61" << std::endl << "68=7+61" << std::endl << "70=3+67" << std::endl << "72=5+67" << std::endl << "74=3+71" << std::endl << "76=3+73" << std::endl << "78=5+73" << std::endl << "80=7+73" << std::endl << "82=3+79" << std::endl << "84=5+79" << std::endl << "86=3+83" << std::endl << "88=5+83" << std::endl << "90=7+83" << std::endl << "92=3+89" << std::endl << "94=5+89" << std::endl << "96=7+89" << std::endl << "98=19+79" << std::endl << "100=3+97";}

en......有亿点点长

#include <bits/stdc++.h>
using namespace std;
int main()
{
    cout << "6=3+3" << endl << "8=3+5" << endl << "10=3+7" << endl << "12=5+7" << endl << "14=3+11" << endl << "16=3+13" << endl << "18=5+13" << endl << "20=3+17" << endl << "22=3+19" << endl << "24=5+19" << endl << "26=3+23" << endl << "28=5+23" << endl << "30=7+23" << endl << "32=3+29" << endl << "34=3+31" << endl << "36=5+31" << endl << "38=7+31" << endl << "40=3+37" << endl << "42=5+37" << endl << "44=3+41" << endl << "46=3+43" << endl << "48=5+43" << endl << "50=3+47" << endl << "52=5+47" << endl << "54=7+47" << endl << "56=3+53" << endl << "58=5+53" << endl << "60=7+53" << endl << "62=3+59" << endl << "64=3+61" << endl << "66=5+61" << endl << "68=7+61" << endl << "70=3+67" << endl << "72=5+67" << endl << "74=3+71" << endl << "76=3+73" << endl << "78=5+73" << endl << "80=7+73" << endl << "82=3+79" << endl << "84=5+79" << endl << "86=3+83" << endl << "88=5+83" << endl << "90=7+83" << endl << "92=3+89" << endl << "94=5+89" << endl << "96=7+89" << endl << "98=19+79" << endl << "100=3+97";
}

#include<iostream>
using namespace std;
bool check(int n)
{
    if(n==0 || n==1)
    {
        return false;
    }
    for(int i=2;i<n;i++)
    {
        if (n%i==0)
        {
            return false;
        }
    }
    return true;
}
int main()
{
    int sum,ans=0;
    cin >> sum;
    for(int i=sum;i>=0;i--)
    {
        if (check(i) && check(sum-i))
        {
            if (i*(sum-i)>=ans)
            {
                ans=i*(sum-i);
            }
        }
    }
    cout << ans;
    return 0;
}

解析

将6~100中的每个数表示成两个素数相加

枚举其中的一个加数，判断两个加数是否都为素数

题解

#include <iostream>
using namespace std;
bool isP(int x)
{
    if (x < 2)
        return false;
    for (int i = 2; i * i <= x; i++)
        if (x % i == 0)
            return false;
    return true;
}
int main()
{
    for (int i = 6; i <= 100; i += 2)
    {
        //找到 i 拆分的方案
        for (int j = 2; j <= i / 2; j++)//现在的方案是 j+(i-j) == i
            if (isP(j) && isP(i - j))
            {
                cout << i << "=" << j << "+" << i - j << "\n";
                break;
            }
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
bool jntm(int n)
{
    if (n<2) return false; 
    for (int i=2;i*i<=n;i++)
    {
        if (n%i==0) return false;
    }
    return true;
}
int main()
{
    for (int j=6;j<=100;j+=2)
    {
        for (int k=2;k<=j;k++)
        {
            int f=j-k;
            if (jntm(k) && jntm(f)) 
            {
                cout<<j<<"="<<k<<"+"<<f<<endl; 
                break;
            }
        }
    }
    return 0;
}

思路：

既然是要把6到100之间的所有偶数拆开，那可以先写个for循环把6到100之间的所有偶数全枚举出来（循环变量为i）

然后，再写个for循环枚举2到i之间所有的数（循环变量为j），通过i-j算出另一个数，如果两个数都是质数，那么输出，然后用个break跳出第二层循环（因为我们只需要把第一种结果输出出来就行了）

#include <iostream>
using namespace std;

bool prime(int n)
{
    for(int i = 2;i < n;i++)
    {
        if(n % i == 0)
            return false;
    }
    return true;
}

int main()
{
    for(int i = 6;i <= 100;i+=2)
    {
        for(int j = 2;j <= i - 2;j++)
        {
            if(prime(j) && prime(i - j))
            {
                cout << i << '=' << j << '+' << i - j << endl;
                break;
            }
        }
    }
    return 0;
}

#include<iostream>
using namespace std;
int main()
{
    int sushu[] = {0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97};
    bool flag = false;
    for(int i = 6; i <= 100; i += 2)
    {
        for(int j = 1; j <= 25; j++)
        {
            for(int k = j; k <= 25; k++)
            {
                if(sushu[j] + sushu[k] == i)
                {
                    cout << i << "=" << sushu[j] << "+" << sushu[k] << endl;
                    flag = true;
                    break;
                }
            }
            if(flag)
            {
                flag = false;
                break;
            }
        }
    }
    return 0;
}