分析

时间复杂度

状态数量 * 状态转移

$O(n * k)$

c++代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 110, inf = 0x3f3f3f3f;

int n, k;
int a[N];
int f[N][N];

int main()
{
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];

    memset(f, -0x3f, sizeof(f));
    f[0][0] = 0;
    for (int i = 1; i <= n; i++)
        for (int j = 0; j < k; j++)
        {
            // f[i][j] = max(f[i - 1][j], f[i - 1][(j + k - a[i] % k) % k] + a[i]);
            f[i][j] = max(f[i][j], f[i - 1][j]);
            f[i][j] = max(f[i][j], f[i - 1][(j + k - a[i] % k) % k] + a[i]);
        }

    cout << f[n][0] << endl;

    return 0;
}

参考文献

闫氏Dp分析法😂
AcWing蓝桥杯