@[Gcysnd](/user/928434)
```cpp
#include <bits/stdc++.h>
using namespace std;
bool one(int x){
for (int i=2;i<=sqrt(x);i++){
if (x%i==0){
return false;
}
}
return true;
}
int main()
{
int a;
cin>>a;
for (int i=1;i<=a/2-1;i++){
int b=2*i+2;
for (int g=2;g<b;g++){
if (one(g)&&one(b-g)){
cout<<b<<"="<<g<<"+"<<b-g<<endl;
break;
}
}
}
return 0;
}
```
by love20110429 @ 2024-02-26 20:32:02
@[Gcysnd](/user/928434) 求关
by love20110429 @ 2024-02-26 20:32:29
@[Gcysnd](/user/928434) 可以发现对于 $(i,j)$ 而言有唯一的 $k$ 使得 $j+k=i$ 所以可以直接计算 $k=j-i$，就不用循环遍历 $k$ 了
by zhouzihang1 @ 2024-02-26 20:35:34
@[zhouzihang1](/user/827018) 谢谢，已解决
by Gcysnd @ 2024-02-26 20:41:41