Number selection (greed)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
#define int long long
void solve() {
	int n;
	cin >> n;
	vector<pii> a(n);
	int cnt = 0;
	for (int i = 0 ;i < n ;i ++) {
		int t;
		cin >> t;
		cnt += t;
		a[i] = mp(t, i);
	}	
	sort(all(a));
	int s = n;
	vector<int> ans;
	bool f=  0;
	for (int s = (cnt / n) * n; s ; s -= n) {
		int t = s;
		bool f = 0;
		ans.clear();
		for (int i = n - 1; i >= 0; i --) {
			if (s - a[i].fi >= 0) {
				s -= a[i].fi;
				ans.pb(a[i].se + 1);
			}

			if (s == 0) {
				f = 1;
				break;
			}
		}	
		if (f) {
			sort(all(ans));
			cout << ans.size() << endl;
			for (auto t : ans) cout << t << " ";
			return;
		}
		else s  = t;
	}
	puts("-1");
}
signed main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

Choose from large to small . For every kind of N, Let's see if we can get 0, If you can output the answer now . Otherwise final output -1