Codeforces Round #267 (Div. 2) C. George and Job（DP）补题

　　　　　　　　　　　　　　　　　　　　　　　　Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~

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 integersp₁, 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

 #include <iostream>
 #include <cstdio>
 #define LL long long
 using namespace std;
 const int maxn =  + ;
 LL p[maxn],s[maxn],dp[maxn][maxn];
 int main(){
     int n,m,k;
     cin>>n>>m>>k;
     for(int i = ;i <= n;i++){
         cin>>p[i];s[i] = s[i - ] + p[i];
     }
     for(int i = m;i <= n;i++)
         for(int j = ;j <= k;j++)
                 dp[i][j] = max(dp[i-m][j-]+s[i]-s[i-m],dp[i-][j]);
     cout<<dp[n][k]<<endl;
     return ;
 }