最大子序列模和(Maximum Subsequence) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    long long m;
    if (!(cin >> n >> m)) return 0;
    vector<long long> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        a[i] %= m;
    }
    int n1 = n / 2;
    int n2 = n - n1;
    vector<long long> L(1 << n1, 0), R(1 << n2, 0);
    for (int i = 0; i < n1; ++i) {
        int bit = 1 << i;
        for (int mask = 0; mask < bit; ++mask) {
            L[mask | bit] = (L[mask] + a[i]) % m;
        }
    }
    for (int i = 0; i < n2; ++i) {
        int bit = 1 << i;
        for (int mask = 0; mask < bit; ++mask) {
            R[mask | bit] = (R[mask] + a[n1 + i]) % m;
        }
    }
    sort(R.begin(), R.end());
    R.erase(unique(R.begin(), R.end()), R.end());
    long long ans = 0;
    for (long long x : L) {
        ans = max(ans, (x + R.back()) % m);
        auto it = lower_bound(R.begin(), R.end(), m - x);
        if (it != R.begin()) {
            --it;
            ans = max(ans, x + *it);
        }
    }
    cout << ans << "\n";
    return 0;
}