Jeff and Digits_jeff's got n cards, each card contains either digi-优快云博客

本文链接：https://blog.youkuaiyun.com/geekdoe/article/details/50499347

本文探讨了如何通过一系列仅包含数字0和5的卡片构建最大的可以被90整除的数字。文章提供了详细的解析思路及C++实现代码，旨在帮助读者理解如何有效地解决这一数学编程问题。

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Description

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 Input

Input

4
5 0 5 0

Output

Input

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

Output

5555555550

Hint

In the first test you can make only one number that is a multiple of 90 — 0.

In the second test you can make number 5555555550, it is a multiple of 90.

先用5找到能被9整除的数，后面加上所有的0。

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

int main(){
    int n,a,i,j,num5,num0,r;
    while(cin>>n){
        num5 = 0;
        num0 = 0;
        for(i = 0; i<n; i++){
            cin>>a;
            if(a == 5){
                num5++;
            }
            else if(a == 0){
                num0++;
            }
        }
        r = num5/9;
        if(!num0){
            cout<<"-1"<<endl;
        }
        else{
            if(r && num0){
                for(i = 0; i<r*9; i++){
                    cout<<"5";
                }
                for(i = 0; i<num0; i++){
                    cout<<"0";
                }
            }
            else{
                cout<<"0";
            }
            cout<<""<<endl;
        }
    }
    return 0;
}