[CF467C] George and Job

The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn’t have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.

Given a sequence of n integers p₁, p₂, …, p_n. You are to choose k pairs of integers:

[l₁, r₁], [l₂, r₂], …, [l_k, r_k] (1 ≤ l₁ ≤ r₁ < l₂ ≤ r₂ < … < l_k ≤ r_k ≤ n; r_i - l_i + 1 = m),  in such a way that the value of sum is maximal possible. Help George to cope with the task.

Input

The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p₁, p₂, …, p_n (0 ≤ p_i ≤ 10⁹).

Output

Print an integer in a single line — the maximum possible value of sum.

Sample test(s)

input

5 2 1 1 2 3 4 5

1 2	5 2 1 1 2 3 4 5

output

 

9

1

9

input

7 1 3 2 10 7 18 5 33 0

1 2	7 1 3 2 10 7 18 5 33 0

output

 

61

1

61

---

题目要求在$n$个数中选择$m$组长度为$k$的子序列使得和最大。

很容易想到，设 $f[i][j]$ 表示选择了i组，并且最后一组以j结尾的最大值。

那么 $\large f[i][j]= max_{p\leq j-k+1} \left ( f[i-1][p] + sum[i]-sum[i-k] \right )$。

这显然是$O(N^3)$的过不了。

所以用$g[p]$表示$f[i-1][1\rightarrow p]$的最大值。

这样就可以$O(1)$转移了。

---

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
#define reg register 
inline char gc() {
    static const int BS = 1 << 22;
    static unsigned char Buf[BS], *st, *ed;
    if (st == ed) ed = Buf + fread(st = Buf, 1, BS, stdin);
    return st == ed ? EOF : *st++;
}
#define gc getchar
inline int read() {
    int res=0;char ch=getchar();bool fu=0;
    while(!isdigit(ch)) {if(ch=='-')fu=1;ch=getchar();}
    while(isdigit(ch))res=(res<<3)+(res<<1)+(ch^48), ch=getchar();
    return fu?-res:res;
}
#define ll long long 

int n, m, k;
ll f[5005][5005], g[5005];
ll sum[5005];
ll ans;

int main()
{
    n = read(), k = read(), m = read();
    for (reg int i = 1 ; i <= n ; i ++) sum[i] = sum[i-1] + read();
    for (reg int i = k ; i <= n ; i ++) f[1][i] = sum[i] - sum[i-k], g[i] = max(g[i-1], f[1][i]);
    for (reg int i = 2 ; i <= m ; i ++)
    {
        for (reg int j = k ; j <= n ; j ++)
            f[i][j] = g[j-k] + sum[j] - sum[j-k];
        memset(g, 0, sizeof g);
        for (reg int j = 1 ; j <= n ; j ++) 
            g[j] = max(g[j-1], f[i][j]);
    }
    for (reg int i = k ; i <= n ; i ++) ans = max(ans, f[m][i]);
    cout << ans << endl;
    return 0;
}