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A. Jeff and Digits(cf)

程序员朝龙技术 2022年11月23日

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A. Jeff and Digitstime limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Jeff’s got n cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he’s got?

Jeff must make the number without leading zero. At that, we assume that number 0 doesn’t contain any leading zeroes. Jeff doesn’t have to use all the cards.

Input

The first line contains integer n (1 ≤ n ≤ 10³). The next line contains n integers a₁, a₂, …, a_n (a_i = 0 or a_i = 5). Number a_i represents the digit that is written on the i-th card.

Output

In a single line print the answer to the problem — the maximum number, divisible by 90. If you can’t make any divisible by 90 number from the cards, print -1.

Sample test(s)Input

4
5 0 5 0

Output

Input

11
5 5 5 5 5 5 5 5 0 5 5

Output

5555555550

 #include <cstdio>
 #include <iostream>
 using namespace std;
 int main()
 {
     int n,x;
     while(~scanf("%d",&n))
     {
         int cnt5 = ,cnt0 = ,max = ;
         while(n--)
         {
             scanf("%d",&x);
             if (x==)
                 cnt5++;
             else
                 cnt0++;
             if(cnt5%==&&cnt5)//能被9整除的数其各位数之和是9的倍数，故5的个数应是9的倍数
             {
                 if (cnt5 > max)
                     max = cnt5;
             }
         }
         if (cnt0==)//能被90整除的数末尾必含0
             puts("-1");
         else if (!max)//若5的个数小于9个且cnt0!=0，则最大的数为0
             puts("");
         else
         {
             while(max--)
                 cout<<"";
             while(cnt0--)
                 cout<<"";
             puts("");
         }     }
     return ;
 }