题目描述：
Your friend has recently learned about coprime numbers. A pair of numbers {a, b} is called coprime if the maximum number that divides both a and b is equal to one.

Your friend often comes up with different statements. He has recently supposed that if the pair (a, b) is coprime and the pair (b, c) is coprime, then the pair (a, c) is coprime.

You want to find a counterexample for your friend’s statement. Therefore, your task is to find three distinct numbers (a, b, c), for which the statement is false, and the numbers meet the condition l ≤ a < b < c ≤ r.

More specifically, you need to find three numbers (a, b, c), such that l ≤ a < b < c ≤ r, pairs (a, b) and (b, c) are coprime, and pair (a, c) is not coprime.

input：
The single line contains two positive space-separated integers l, r (1 ≤ l ≤ r ≤ 1018; r - l ≤ 50).

output：
Print three positive space-separated integers a, b, c — three distinct numbers (a, b, c) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.

If the counterexample does not exist, print the single number -1.

样例输入：
2 4
样例输出：
2 3 4
样例输入：
10 11
样例输出：
-1
样例输入：
900000000000000009 900000000000000029
样例输出：
900000000000000009 900000000000000010 900000000000000021

题目大意：
我们要在r,l期间内找出三个数，这三个数要满足（a,b）互为质数，（b,c）互为质数，并且（a,c）不互为质数，如果在r,l期间有多组这样的数，随意输一组，按照小到大的顺序输出，如果没有则输出-1；
注：互为质数即两数的最大公因数是1
这个题就是典型的gcd题，并且虽然他的数据达到1e18但是，我们发现r,l之间的差值3不大于50，因此我们完全可以用三层for循环来进行查找；话不多说，上代码：

#include <iostream>
#include <cstdio>

using namespace std;
#define ll long long

ll gdc(ll a,ll b) { //找到两数的最大公因数
	while(b%a != 0){
		int ans = b%a;
		b = a;
		a = ans;
	}
	return a;
}
int main () {
	ll r,l,i,j,k;
	bool f = 0;
	cin >> r >> l;
	for (i = r;i <= l;i++) {
		for (j = i+1;j <= l;j++) {
			for (k = j+1;k <= l;k++) {
				if(gdc(i,j)==1&&gdc(j,k)==1&&gdc(i,k)!=1){//满足条件直接输出，然后跳出循环
					f = 1;
					cout << i << " " << j << " " << k << "\n";
					break;
				}
			}
			if(f)
				break;
		}
		if(f)
			break;
	}
	if(!f)
		cout << "-1\n";
	return 0;
}