题目链接

题意

让你构造出一个Codeforces 743C. Vladik and fractions(数学+思维)_#define 给出你n让你输出满足条件的Codeforces 743C. Vladik and fractions(数学+思维)_#include_02

题解

首先我们看到n的分子上有一个2,那么我们可以直接拿出来一个,让a=n,那就剩Codeforces 743C. Vladik and fractions(数学+思维)_#include_03,然后我们呢利用Codeforces 743C. Vladik and fractions(数学+思维)_#define_04. 我们通分后面那两个时,会把(n+1)消掉.当然我们需要特判n=1的时候,这是误解的.

#include <set>
#include <map>
#include <queue>
#include <string>
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;
typedef unsigned long long ull;

#define
#define
#define
#define
#define
#define
#define
#define
#define
#define

const int MOD = 10007;
const int mod = 998244353;
const int maxn = 2e5 + 1010;
const int dx[] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy[] = {1, 0, -1, -1, 1, -1, 1, 0};

ll a[maxn], b[maxn];
ll ans, n, m, len, k;
string s;
void solve()
{
cin >> n;
if(n==1) cout<<"-1"<<endl;
else
cout<<n<<" "<<n+1<<" "<<n*(n+1)<<endl;
}

int main()
{
solve();
return 0;
}