#include #include typedef long long int ll; typedef long long int ull; #define MP make_pair using namespace std; using namespace atcoder; typedef pair P; // const ll MOD = 998244353; const ll MOD = 1000000007; // using mint = modint998244353; using mint = modint1000000007; const double pi = 3.1415926536; const int MAX = 2000003; long long fac[MAX], finv[MAX], inv[MAX]; void COMinit() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; for (int i = 2; i < MAX; i++){ fac[i] = fac[i - 1] * i % MOD; inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD; finv[i] = finv[i - 1] * inv[i] % MOD; } } // 二項係数計算 long long COM(int n, int k){ if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD; } ll gcd(ll x, ll y) { if (y == 0) return x; else if (y > x) { return gcd (y, x); } else return gcd(x % y, y); } ll lcm (ll x, ll y) { return x / gcd(x, y) * y; } ll my_sqrt(ll x) { ll m = 0; ll M = 3000000001; while (M - m > 1) { ll now = (M + m) / 2; if (now * now <= x) { m = now; } else { M = now; } } return m; } ll keta(ll n) { ll ret = 0; while (n) { n /= 10; ret++; } return ret; } ll ceil(ll n, ll m) { // n > 0, m > 0 ll ret = n / m; if (n % m) ret++; return ret; } ll pow_ll(ll x, ll n) { if (n == 0) return 1; if (n % 2) { return pow_ll(x, n - 1) * x; } else { ll tmp = pow_ll(x, n / 2); return tmp * tmp; } } ll dp[5001][5001]; ll INF = 100000000000; int main() { int n; cin >> n; int a[n]; for (int i = 0; i < n; i++) cin >> a[i]; int now0 = 0; int now1 = 0; vector

v; int all_z = 0; for (int i = 0; i < n; i++) { if (a[i] == 2) { v.push_back(P(now0, now1)); now0 = 0; now1 = 0; } else if (a[i] == 0) { now0++; all_z++; } else { now1++; } } v.push_back(P(now0, now1)); int r = v.size(); for (int i = 0; i <= 5000; i++) { for (int j = 0; j <= 5000; j++) { dp[i][j] = INF; } } dp[0][0] = 0; for (int i = 0; i < r; i++) { for (int j = 0; j <= 5000; j++) { // 0埋めする場合 if (j + v[i].first + v[i].second <= 5000) { dp[i + 1][j + v[i].first + v[i].second] = min(dp[i + 1][j + v[i].first + v[i].second], dp[i][j] + v[i].second); } // 1埋めする場合 dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]); } } if (dp[r][all_z] == INF) cout << -1 << endl; else cout << dp[r][all_z] << endl; return 0; }