problem link

Since $n\le 21$, $\mathcal{O}(n^3)$ brute force enumeration would suffice. Lexicographic order are trivial in this case with nested for loops.

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
int n;
inline int read()
{int s=0,w=1;char ch=getchar();while(ch<'0' || ch>'9'){if(ch=='-')w-=1;ch=getchar();}while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();return s*w;
}
int main()
{n=read();for(int i=0;i<=n;++i)for(int j=0;j<=n;++j)for(int k=0;k<=n;++k)if(i+j+k<=n){printf("%d %d %d\n",i,j,k);}return 0;
}