据原贴的原理随手造的，期望能卡过大部分正向错解。

#include <bits/stdc++.h>
using namespace std;

int n,m,k;

int main(){
	mt19937 rng(time(0));
	n=rng()%9996+5;
	m=n+(n-1)/2-1,k=rng()%100+1;
	cout<<n<<' '<<m<<" "<<k<<"\n";
	for(int i=1;i<n;i++){
		cout<<i<<" "<<i+1<<" "<<rng()/k%(4*k)+rng()%k<<'\n';
	}
	for(int i=1;i<(n-1)/2+1;i++){
		cout<<i<<" "<<n-i<<" "<<n-i+rng()%500<<'\n';
	}
	return 0;
}