CF#204DIV2:A. Jeff and Digits

http://codeforces.com/contest/352/problem/A

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,
..., an (ai = 0 or
ai = 5). Number
ai 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

0

Input

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

Output

5555555550

Note
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.



水题，找出所有数能组成的最大的能被90整除的数

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int a;
int main()
{
    int n,i,j,num5,num0,r;
    while(~scanf("%d",&n))
    {
        num5 = 0;
        num0 = 0;
        for(i = 0; i<n; i++)
        {
            scanf("%d",&a);
            if(a == 5)
                num5++;
            else if(a == 0)
                num0++;
        }
        r = num5/9;//9的倍数个5组合才能被9整除
        if(!num0)//必须含有0            printf("-1\n");
        else
        {
            if(r && num0)
            {
                for(i = 0; i<r*9; i++)
                    printf("5");
                for(i = 0; i<num0; i++)
                    printf("0");
            }
            else
                printf("0");
            printf("\n");
        }
    }

    return 0;
}