Educational Codeforces Round 69 (Rated for Div. 2) D. Yet Another Subarray Problem 【数学+分块】

I, entitled

　　D. Yet Another Subarray Problem

Second, analysis

　　Reference of deriving formulas Los Valley get together their derivation

　　Focus derivation of the formula, the formula is derived, the block is easily thought. But it is easy to write fried.

　　1 easy to overflow in some places, and that the size of infinite value of the set is also relevant.

　　If every 2 to determine the boundaries of $ r $, then in the case of the remainder of enumeration $ m $, must have noticed is greater than $ r $ still can not be enumerated.

Three, AC codes

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 #define Min(a, b) ((a)<(b)?(a):(b))
 5 #define Max(a, b) ((a)>(b)?(a):(b))
 6 typedef long long ll;
 7 const int maxn = 3e5 + 13;
 8 const ll inf = 1e15 ;
 9 int n, m;
10 ll k;
11 int a[maxn];
12 ll sum[maxn];
13 ll D[maxn];
14 ll Dmin[15];
15 
16 int main()
17 {
18     //freopen("input.txt", "r", stdin);
19     while(scanf("%d %d %I64d", &n, &m, &k) != EOF)
20     {
21         fill(Dmin, Dmin + 11, inf);
22         sum[0] = 0;
23         for(int i = 1; i <= n; i++)
24         {
25             scanf("%d", &a[i]);
26             sum[i] = sum[i - 1] + a[i];
27             D[i] = sum[i] - k * (i / m);
28         }
29         ll ans = 0;
30         Dmin[0] = 0;
31         for(int i = 1; i <= n; i++)
32         {
33             ll res = -inf;
34             for(int j = 0; j < m; j++)
35             {
36                 int f = ceil(1.0 * ( (i % m) - j) / m);
37                 res = Max(res, D[i] - Dmin[j] - k * f);
38             }
39             Dmin[i % m] = Min(D[i], Dmin[i % m]);
40             ans = Max(res, ans);
41         }
42         printf("%lld\n", ans);
43     }
44 
45 
46 }