Codeforces Round #313 (Div. 2) A. Currency System in Geraldion 水题

A. Currency System in Geraldion

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/560/problem/A

Description

A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimumunfortunate sum?

Input

The first line contains number n (1 ≤ n ≤ 1000) — the number of values of the banknotes that used in Geraldion.

The second line contains n distinct space-separated numbers a1, a2, ..., an (1 ≤ ai ≤ 106) — the values of the banknotes.

Output

Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print  - 1.

Sample Input

5
1 2 3 4 5

Sample Output

-1

HINT

 

题意

让你输出，用这些钱不能表示的最小的和

如果没有这个数 输出-1

题解：

如果给你的数含有1，就输出-1

如果不含有1，就输出-1……

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=202501;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int main()
{
    int n=read();
    for(int i=0;i<n;i++)
    {
        int x=read();
        if(x==1)
        {
            printf("-1\n");
            return 0;
        }
    }
    printf("1\n");
}