ABC335B - Tetrahedral Number

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
#include
#include
#include
#include
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;
}