方法一：数论分析简单

分析 x%30 == 0 等价于 x % 3 == 0 && x % 10 == 0
如果想要%10 则最后一位为0;
如果想要%3 则各位的数字之和可以%3 ==0；
所以如果想要最大值，那么一定是逆序排列！

#include <bits/stdc++.h>
using namespace std;
#define int long long
//typedef long long ll;
//typedef pair<int, int> PII;
// const double PI = 3.14159265358979;
//const int MOD=998244353
const int N=2e5+7;
const int INF = 0x3f3f3f3f;
//vector<int>ve;

void solve()
{
    string s;
    cin >> s;

    sort(s.begin(), s.end(),greater<>());

    if(stoi(s) % 30 ==0) cout << s <<endl;
    else cout << -1 <<endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);

    int T=1;
    // cin >> T;
    while(T--) solve();
    return 0;
}

方法二：dfs搜索

#include <bits/stdc++.h>
using namespace std;
// #define int long long
// typedef long long ll;
// typedef pair<int, int> PII;
//  const double PI = 3.14159265358979;
// const int MOD=998244353
const int N = 2e5 + 7;
const int INF = 0x3f3f3f3f;
// vector<int>ve;
string s;
int cnt[9];
int res = -1;
int n;

void dfs(int u, int s)
{

    if (u == n && s % 30 == 0)
    {
        res = max(res, s);
        return;
    }
    for (int i = 0; i <= 9; i++)
    {
        if (cnt[i])
        {
            cnt[i]--;
            dfs(u + 1, s * 10 + i);
            cnt[i]++;
        }
    }
}
void solve()
{
    cin >> s;
    n = s.size();
    for (int i = 0; i < s.size(); i++)
        cnt[s[i] - '0']++;

    dfs(0,0);
    cout << res << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int T = 1;
    // cin >> T;
    while (T--)
        solve();
    return 0;
}

方法三：搜索，使用next_permutation()函数

#include <bits/stdc++.h>
using namespace std;
string s;
int n;
int a[9];
int ans = -1;

int main()
{
    cin >> s;
    n = s.size();

    for (int i = 0; i < n; i++)
        a[i] = s[i] - '0';

    sort(a, a + n);

    do
    {
        int res = 0;

        for (int i = 0; i < n; i++)
            res = res * 10 + a[i];

        if (res % 30 == 0)
            ans = max(ans, res);

    } while (next_permutation(a, a + n));
    cout << ans << endl;
    return 0;
}