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：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const ll INF = 0x3f3f3f3f3f3f3f3f;
 5 const int N = 300005;
 6 ll a[N], dp[N][11];
 7 int main()
 8 {
 9     int n; ll m, k;
10     while (cin >> n >> m >> k)
11     {
12         memset(dp, 0, sizeof dp);
13         for (int i = 1; i <= n; i++) cin >> a[i];
14         ll res = 0;
15         dp[0][0] = -k;
16         for (int i = 1; i < m; i++) dp[0][i] = -INF;
17         for (int i = 1; i <= n; i++)
18         {
19             for (int j = 0; j < m; j++)
20             {
21                 if (j == (m == 1 ? 0 : 1))
22                 {
23                     dp[i][j] = max(dp[i - 1][0] + a[i] - k, a[i] - k);
24                 }
25                 else
26                 {
27                     dp[i][j] = max(dp[i - 1][(j - 1 + m) % m] + a[i], a[i] - k);
28                 }
29                 res = max(res, dp[i][j]);
30             }
31         }
32         cout << res << endl;
33     }
34     return 0;
35 }

来源：https://www.cnblogs.com/wangyiming/p/12031533.html