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.
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.
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.
4
5 0 5 0
0
11
5 5 5 5 5 5 5 5 0 5 5
5555555550
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.
%9的性质还记得吗？个位数字相加看%9==0即可；
注意题目是90；简单模拟一下即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int a[maxn];
bool cmp(int a, int b) { return a > b; }
int main() {
//ios::sync_with_stdio(0);
int n; int cnt = 0; cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i]; if (a[i] == 0)cnt++;
}
sort(a + 1, a + 1 + n, cmp);
int sum = 0; int tot = 0; bool fg = 0; int maxx = 0;
for (int i = 1; i <= n; i++) {
if (a[i] == 0)break;
sum += a[i]; tot++;
// cout << sum << endl;
if (sum % 9 == 0) {
fg = 1; maxx = max(maxx, tot);
}
}
if (!fg&&cnt==0)cout << -1 << endl;
else {
if (maxx == 0&&cnt) {
cout << 0 << endl; return 0;
}
if (cnt == 0) {
cout << -1 << endl; return 0;
}
for (int i = 1; i <= maxx; i++)cout << 5;
for (int i = 1; i <= cnt; i++)cout << 0;
}
return 0;
}