dp数组的选择：首先是选不选某个位置是一种状态，且题目中提醒了有余数一项，所以是一个二维数组，dp[i][j]是进行到第i个糖果余数为j的方法总数，且状态转移方程除了原先的这个位置不选外，如果要选就是a[i]+dp[i-1][j+(k-a[i]%k)%k]
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include<climits>
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];

  for (int i = 0; i < 110; ++i)
        for (int j = 0; j < 110; ++j)
            f[i][j] = INT_MIN;
    f[0][0] = 0;
    for (int i = 1; i <= n; i++)
        for (int j = 0; j < k; j++)
        {

            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;
}