Codeforces Round #267 (Div. 2) C. 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

Output

9

Input

7 1 3
2 10 7 18 5 33 0

Output

61
题意：将一个长度为n的序列，分成k段长度为m的子序列，求这k个子序列和的最大值
思路：dp[i][j]表示是前i个数选出j段的最大值，显然有不选这个数，和考虑这个数的两种情况。而考虑这个数的话，因为连续性也只会增加以这个数为结尾的m序列
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn = 5100;

ll num[maxn], sum[maxn], dp[maxn][maxn];
ll n, m, k;

int main() {
	cin >> n >> m >> k;
	for (int i = 1; i <= n; i++) {
		cin >> num[i];
		sum[i] = sum[i-1] + num[i];
	}

	for (int i = m; i <= n; i++)
		for (int j = k; j >= 1; j--)
			dp[i][j] = max(dp[i-1][j], dp[i-m][j-1]+sum[i]-sum[i-m]);

	cout << dp[n][k] << endl;
	return 0;
}