$$ markdown$$
`了解一下`
by Chloris @ 2018-07-08 15:24:17
```cpp
#include<bits/stdc++.h>
using namespace std;
int zt(int x) {
int i;
for(i=2; i<=sqrt(x); ++i)
if(x%i==0)
return 0;
return 1;
}
int a[10001];
int main() {
int n,i,j,k,sum=0,x=0,y;
cin>>n;
for(i=2; i<=n; ++i) {
if(zt(i)==1) a[++sum]=i;
}
for(i=1; i<=sum; ++i)
for(j=1; j<=sum; ++j) {
if(zt(n-a[i]-a[j])==1) {
cout<<a[i]<<" "<<a[j]<<" "<<n-a[i]-a[j];
return 0;
}
}
return 0;
}
```
by Siyuan @ 2018-07-08 15:28:59
@[siyuan](/space/show?uid=49725) 能给一个正解吗？
by xikiy @ 2018-07-08 18:16:45
@[Chloristendika](/space/show?uid=76527) 能具体一点吗？
by xikiy @ 2018-07-08 18:17:00
```
#include<bits/stdc++.h>
using namespace std;
int main()
{
int i,n,j,k,a[20001]={0};
cin>>n;
for(i=2;i<=n;i++)
{
for(k=2;k<=sqrt(i);k++)
if(i%k==0)
break;
if(k>sqrt(i))
a[i]=1;
}
for(i=2;i<=n;i++)
for(k=2;k<=n;k++)
if(a[i]==1&&a[k]==1&&a[n-i-k]==1)
{
cout<<i<<" "<<k<<" "<<n-i-k;
return 0;
}
}
```
by 白烛葵 @ 2018-07-19 08:16:49
水题一个
by 白烛葵 @ 2018-07-19 08:17:18