結果

問題 No.2840 RGB Plates
ユーザー milkcoffee
提出日時 2024-06-22 14:59:26
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,143 bytes
コンパイル時間 1,979 ms
コンパイル使用メモリ 199,680 KB
最終ジャッジ日時 2025-02-22 00:10:43
ジャッジサーバーID
（参考情報） 		judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 17 WA * 11
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
// #if __has_include(<atcoder/all>)
// #include <atcoder/all>
// using namespace atcoder;
// #endif
using namespace std;
// #include "boost/multiprecision/cpp_int.hpp"
// using namespace boost::multiprecision;
// #pragma GCC target("avx2")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")
#define ll long long
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define rep_up(i, a, n) for (ll i = a; i < n; ++i)
#define rep_down(i, a, n) for (ll i = a; i >= n; --i)
#define P pair<ll, ll>
#define pb push_back
#define bit_count(x) __builtin_popcountll(x)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) a / gcd(a, b) * b
// #define endl "\n"
#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define vll vector<ll>
#define vvll vector<vll>
#define vvvll vector<vvll>
#define vvvvll vector<vvvll>
#define pqll priority_queue<ll>
#define pqllg priority_queue<ll, vector<ll>, greater<ll>>
template <class T>
inline void vin(vector<T> &v) { rep(i, v.size()) cin >> v.at(i); }
template <class T>
using V = vector<T>;
constexpr ll INF = (1ll << 60);
// constexpr ll mod = 1000000007;
// constexpr ll mod = 998244353;
// using mint = atcoder::modint;
// using mint = atcoder::modint998244353;
// using mint = atcoder::modint1000000007;
#define vmint vector<mint>
#define vvmint vector<vmint>
#define vvvmint vector<vvmint>
#define vvvvmint vector<vvvmint>
constexpr double pi = 3.14159265358979323846;
random_device seed_gen;
mt19937 engine(seed_gen());
template <typename T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
void pt(T val)
{
    cout << val << "\n";
}
template <typename T>
void pt_vll(vector<T> &v)
{
    ll vs = v.size();
    rep(i, vs)
    {
        cout << v[i];
        if (i == vs - 1)
            cout << "\n";
        else
            cout << " ";
    }
}
template <typename T>
void pt_vmint(vector<T> &v)
{
    ll vs = v.size();
    rep(i, vs)
    {
        cout << v[i].val();
        if (i == vs - 1)
            cout << "\n";
        else
            cout << " ";
    }
}
ll mypow(ll a, ll n)
{
    ll ret = 1;
    if (n == 0)
        return 1;
    if (a == 0)
        return 0;
    rep(i, n)
    {
        if (ret > (ll)(9e18 + 10) / a)
            return -1;
        ret *= a;
    }
    return ret;
}
long long modpow(long long a, long long n, long long mod)
{
    long long res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
long long modinv(long long a, long long m)
{
    long long b = m, u = 1, v = 0;
    while (b)
    {
        long long t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}
void solve();
// ここまでテンプレ
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    // cout << fixed << setprecision(16);
    // ll T;
    // cin >> T;
    // rep(ca, T)
    solve();
}
// WA: K = 5000 とする
void solve()
{
    ll n, m, k, cnt = 0, sum = 0, ans = 0;
    cin >> n;
    assert(3 <= n && n <= 3000);
    vll a;
    ll x;
    while (cin >> x)
    {
        a.push_back(x);
    }
    assert(a.size() == n);
    rep(i, n) assert(1 <= a[i] && a[i] <= 5000);
    m = 5000;
    vvll dp(n + 1, vll(m + 1));
    dp[0][0] = 1;
    rep(i, n)
    {
        ans += a[i];
        rep(j, m + 1)
        {
            ll x = j + a[i];
            dp[i + 1][j] += dp[i][j];
            chmin(dp[i + 1][j], 2ll);
            if (x > m)
                continue;
            dp[i + 1][x] += dp[i][j];
            chmin(dp[i + 1][x], 2ll);
        }
    }
    ll mi = INF;
    rep(i, m)
    {
        if (dp[n][i + 1] == 2)
        {
            chmin(mi, i + 1);
        }
    }
    // pt(mi);
    if (mi == INF || ans <= 2 * mi)
    {
        pt(-1);
        return;
    }
    ans -= 2 * mi;
    pt(ans);
}
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