Codeforces Round #339 (Div. 2) B. Gena's Code 水题

B. Gena's Code

题目连接：

http://www.codeforces.com/contest/614/problem/B

Description

It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!

There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.

Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.

Input

The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.

It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.

Output

Print a single number without leading zeroes — the product of the number of tanks presented by each country.

Sample Input

3

5 10 1

Sample Output

50

Hint

题意

给你n个数，让你输出他们的乘积

保证里面最多只有一个数不是美丽数

美丽数定义：这个数在10进制下，只含有0和1，并且最多只有一个1

题解：

如果其中有0，那么答案肯定是0

首先我们找到不是美丽数的那个数num，那么答案肯定是num*10^k

这个k就是每一个美丽数的0的个数的和就好了

代码

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

string s;
string ans;
string tmp;
int num = 0;
int main()
{
    int n;
    scanf("%d",&n);
    int flag = 0;
    for(int i=0;i<n;i++)
    {
        cin>>tmp;
        if(tmp=="0")
        {
            cout<<"0"<<endl;
            return 0;
        }
        int cnt = 0;
        int flag = 0;
        for(int j=0;j<tmp.size();j++)
        {
            if(tmp[j]=='1')
                cnt++;
            if(tmp[j]!='1'&&tmp[j]!='0')
                flag = 1;
        }
        if(flag||cnt>1)
            ans = tmp;
        else
            num += tmp.size()-1;
    }
    if(ans.size())
        cout<<ans;
    else
        cout<<"1";
    for(int i=0;i<num;i++)
        cout<<"0";
    cout<<endl;
}