AtCoder-3856

An adult game master and N children are playing a game on an ice rink. The game consists of K rounds. In the i-th round, the game master announces:

• Form groups consisting of A_i children each!

Then the children who are still in the game form as many groups of A_i children as possible. One child may belong to at most one group. Those who are left without a group leave the game. The others proceed to the next round. Note that it's possible that nobody leaves the game in some round.

In the end, after the K-th round, there are exactly two children left, and they are declared the winners.

You have heard the values of A₁, A₂, ..., A_K. You don't know N, but you want to estimate it.

Find the smallest and the largest possible number of children in the game before the start, or determine that no valid values of N exist.

Constraints

• 1≤K≤10⁵
• 2≤A_i≤10⁹
• All input values are integers.

Input

Input is given from Standard Input in the following format:

K
A1 A2 … AK

Output

Print two integers representing the smallest and the largest possible value of N, respectively, or a single integer −1 if the described situation is impossible.

Sample Input 1

4
3 4 3 2

Sample Output 1

6 8

For example, if the game starts with 6 children, then it proceeds as follows:

• In the first round, 6 children form 2 groups of 3 children, and nobody leaves the game.
• In the second round, 6 children form 1 group of 4 children, and 2 children leave the game.
• In the third round, 4 children form 1 group of 3 children, and 1 child leaves the game.
• In the fourth round, 3 children form 1 group of 2 children, and 1 child leaves the game.

The last 2 children are declared the winners.

Sample Input 2

5
3 4 100 3 2

Sample Output 2

-1

This situation is impossible. In particular, if the game starts with less than 100children, everyone leaves after the third round.

Sample Input 3

10
2 2 2 2 2 2 2 2 2 2

Sample Output 3

2 3

题解：这道题应该倒过来反推；代码如下：

AC代码为：

#include <iostream>  
#include <cstdio>  
using namespace std;


int a[100005];
int main() 
{
int k;
cin >> k;
for (int i = 1; i <= k; i++) 
{
cin >> a[i];
}
long long mmax = 2, mmin = 2;
for (int i = k; i >= 1 && mmax >= mmin; i--)
{
if (mmin%a[i] != 0)
mmin = mmin / a[i] * a[i] + a[i];
mmax = (mmax / a[i] + 1)*a[i] - 1;
}
if (mmax >= mmin)
cout << mmin << ' ' << mmax << endl;
else
cout << -1 << endl;

return 0;
}