CF1197D Yet Another Subarray Problem

思路：

使用动态规划，在经典的最大子段和解法基础上进行扩展。dp[i][j]表示以第i个数为结尾，并且长度模m等于j的所有子段的最大cost。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll INF = 0x3f3f3f3f3f3f3f3f;
 const int N = ;
 ll a[N], dp[N][];
 int main()
 {
     int n; ll m, k;
     while (cin >> n >> m >> k)
     {
         memset(dp, , sizeof dp);
         for (int i = ; i <= n; i++) cin >> a[i];
         ll res = ;
         dp[][] = -k;
         for (int i = ; i < m; i++) dp[][i] = -INF;
         for (int i = ; i <= n; i++)
         {
             for (int j = ; j < m; j++)
             {
                 if (j == (m ==  ?  : ))
                 {
                     dp[i][j] = max(dp[i - ][] + a[i] - k, a[i] - k);
                 }
                 else
                 {
                     dp[i][j] = max(dp[i - ][(j -  + m) % m] + a[i], a[i] - k);
                 }
                 res = max(res, dp[i][j]);
             }
         }
         cout << res << endl;
     }
     return ;
 }