#916532 (C++17) No.2488 Mod Sum Maximization

• 問題: No.2488 Mod Sum Maximization
• ユーザー: square1001
• 提出日時: 2023-09-29 21:16:06
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 638 ms / 2,000 ms
• コード長: 1,469 bytes
• コンパイル時間: 2,145 ms
• コンパイル使用メモリ: 198,268 KB
• 最終ジャッジ日時: 2025-02-17 03:04:49
• ジャッジサーバーID （参考情報）: judge3 / judge1

ファイルパターン	結果
sample	AC * 3
other	AC * 38

ソースコード

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#include <bits/stdc++.h>
using namespace std;
const long long INF = (3LL << 59);
class segtree {
private:
    int sz;
    vector<long long> val;
public:
    segtree() : sz(0), val(vector<long long>()) {}
    segtree(int n) : sz(n >= 3 ? 1 << (32 - __builtin_clz(n)) : n), val(vector<long long>(sz * 2, -INF)) {}
    void update(int pos, long long x) {
        pos += sz;
        val[pos] = x;
        while (pos > 1) {
            pos /= 2;
            val[pos] = max(val[pos * 2], val[pos * 2 + 1]);
        }
    }
    long long rangemax(int l, int r) const {
        l += sz;
        r += sz;
        long long ans = -INF;
        while (l != r) {
            if (l & 1) ans = max(ans, val[l++]);
            if (r & 1) ans = max(ans, val[--r]);
            l /= 2;
            r /= 2;
        }
        return ans;
    }
};
int main() {
    // step #1. input
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vector<int> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    // step #2. calculate sum
    long long suma = 0;
    for (int i = 0; i < N; i++) {
        suma += A[i];
    }
    // step #3. dynamic programming
    vector<long long> dp(N, -INF);
    segtree seg(N);
    dp[N - 1] = 0;
    seg.update(N - 1, dp[N - 1]);
    for (int i = N - 2; i >= 0; i--) {
        int posl = i + 1;
        for (int j = A[i]; j <= A[N - 1]; j += A[i]) {
            int posr = lower_bound(A.begin(), A.end(), j + A[i]) - A.begin();
            long long res = seg.rangemax(posl, posr);
            dp[i] = max(dp[i], res - j);
            posl = posr;
        }
        seg.update(i, dp[i]);
    }
    // step #4. output
    long long ans = suma + dp[0];
    cout << ans << endl;
    return 0;
}
 
 
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