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

Input

7 1 3

2 10 7 18 5 33 0

Output

 #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 ;

 }

巴特西

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

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