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 11 2 3 4 5

output

9

input

7 1 32 10 7 18 5 33 0

output

61

就是k组数，每组是连续的m个数，求这些数最大和。群里有人推荐这道题，于是来写了一下，题目意思还是很好理解的，很简单的动规思路。 连续m个数，那就对每一个位置都算出前m个数的和，然后就是取k次的一个背包。

dp[i][j]=max(dp[i][j-1],dp[i-1][j-m]+d[j]);

i代表取的次数，j代表到第j个数。但是初始化还有一点细节要处理。

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int a[5005];long long dp[5005], dp1[5005][5005];long long max(long long a, long long b){return a > b ? a : b;}int main(){int i, j, m, n, ans, k;long long sum = 0, pre;cin >> n >> m >> k;for (i = 1; i <= n; i++)cin >> a[i];for (i = 0; i < m; i++){dp[i] = 0;sum = sum + a[i];}a[0] = 0; j = 0;for (i = m; i <= n; i++){sum = sum - a[j];sum = sum + a[i];dp[i] = sum;j++;}for (i = 0; i < m; i++)dp1[1][i] = 0;for (i = m; i <= n; i++)dp1[1][i] = max(dp1[1][i - 1], dp[i]);for (i = 2; i <= k; i++){for (j = i * m; j <= n; j++){dp1[i][j] = max(dp1[i][j - 1], dp1[i-1][j - m] + dp[j]);}}//for (i = 1; i <= n; i++)//cout << dp[i] << " ";//cout << endl;//for (i = 1; i <= k; i++)//{//for (j = 1; j <= n; j++)//cout << dp1[i][j] << " ";//cout << endl;//}cout << dp1[k][n] << endl;return 0;}

好久不写动规，这次感觉不错，得写专题了，模拟已经够了。