#867986 (C++14) No.2196 Pair Bonus

• 問題: No.2196 Pair Bonus
• ユーザー: 佐藤竜也
• 提出日時: 2023-05-11 12:47:59
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 228 ms / 2,000 ms
• コード長: 2,290 bytes
• コンパイル時間: 1,520 ms
• コンパイル使用メモリ: 168,296 KB
• 実行使用メモリ: 5,760 KB
• 最終ジャッジ日時: 2024-11-27 11:37:03
• 合計ジャッジ時間: 5,008 ms
• ジャッジサーバーID （参考情報）: judge3 / judge4

ファイルパターン	結果
sample	AC * 3
other	AC * 15

ソースコード

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, s, n) for (int i = (s); i < (int)(n); i++)
#define rep2(i, n, s) for (int i = (n - 1); i >= (s); i--)
#define all(a)  (a).begin(),(a).end()
#define all_c(a, b) (a).begin(), (a).end(), back_inserter((b))
vector<int> dy = {-1, 0, 1, 0, -1, -1, 1, 1};
vector<int> dx = {0, 1, 0, -1, -1, 1, 1, -1};
const ll INF = numeric_limits<long long>::max();
const ll MOD = 1'000'000'007;
ll unused = INF % MOD;
ll gcd(ll A, ll B) {
    if (B == 0) return A;
    return gcd(B, A % B);
}
ll lcm(ll A, ll B) {
    ll g = gcd(A, B);
    return A / g * B;
}
// 二分検索の雛型
bool binary_search(int N, int A[], int K) {
    int left = 0, right = N - 1;
    while (left <= right) {
        int mid = (left + right) / 2;
        if (A[mid] == K) return true;
        if (A[mid] < K) left = mid + 1;
        else right = mid - 1;
    }
    return false;
}
/*
lower_bound -> if (A[mid] < K)
upper_bound -> if (A[mid] <= K)
*/
int binary_search2(int N, ll A[], ll K) {
    int left = 0, right = N;
    while (left < right) {
        int mid = (left + right) / 2;
        if (A[mid] < K) left = mid + 1;
        else right = mid;
    }
    return right;
}
/* メモ帳
・小数点以下表示 / cout << fixed << setprecision(12) << 
・ルートの計算   / sqrt(A)
・順列全列挙
    do {
        
    } while (next_permutation(V.begin() V.end()));
・bit検索
for (int i = 0; i < (1 << N); i++) {
    rep(j, 0, N) {
        int wari = (1 << j);
        if ((i / wari) % 2 == 1) {
            
        }
    }
}
*/
int N, A[200009], B[200009], X[100009], Y[100009];
ll ans = 0;
int main() {
    cin >> N;
    rep(i, 0, 2 * N) cin >> A[i];
    rep(i, 0, 2 * N) cin >> B[i];
    rep(i, 0, N) cin >> X[i];
    rep(i, 0, N) cin >> Y[i];
    
    rep(i, 0, N) {
        ll tmp = 0;
        rep(j, 0, 2) {
            rep(k, 0, 2) {
                ll t = 0;
                if (j == 0) t += A[2 * i];
                else t += B[2 * i];
                if (k == 0) t += A[2 * i + 1];
                else t += B[2 * i + 1];
                if (j == k) t += X[i];
                else t += Y[i];
                tmp = max(tmp, t);
            }
        }
        ans += tmp;
    }
    
    cout << ans << endl;
}
 
 
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