Codeforces 467C George and Job (DP)

C. George and Job

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., pn. You are to choose k pairs of integers:

 

[l1, r1], [l2, r2], ..., [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < ... < lk ≤ rk ≤ n; ri - li + 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, ..., pn(0 ≤ pi ≤ 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：从一个长度为n的数列中选出k段区间长度为m的段，使得这k段数的和最大。

思路：DP方程：dp[i][j] = max(dp[i - 1][j], dp[i - m][j - 1] + sum[i] - sum[i - m]);

　　　dp[i][j]表示前i个数中选j段的最大和，sum[i]表示前i个数的和。

#include<algorithm>
#include<iostream>
#include<limits.h>
#include<stdlib.h>
#include<string.h>
#include<complex>
#include<cstring>
#include<iomanip>
#include<stdio.h>
#include<bitset>
#include<cctype>
#include<math.h>
#include<string>
#include<time.h>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>

#define LL long long

using namespace std;
const LL mod = 1e9 + 7;
const double PI = acos(-1.0);
const double E = exp(1.0);
const int M = 5005;

int a[M];
LL ans[M];
LL dp[M][M];

int main()
{
    int n, m, k;
    cin >> n >> m >> k;
    for(int i = 1; i <= n; ++i){
        cin >> a[i];
        ans[i] = ans[i - 1] + a[i];
    }
    for(int i = m; i <= n; ++i)
        for(int j = 1; j <= k; ++j)
            dp[i][j] = max(dp[i - 1][j],dp[i - m][j - 1] + ans[i] - ans[i - m]);
    cout << dp[n][k] << endl;
    return 0;
}