Codeforces Round #267 (Div. 2)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

给n个数，问把这个数列分为k段长度为m的区间的元素之和的最大值是多少。

dp[i][j]由dp[i - 1][j]不添加区间或者由dp[i - m][j - 1]添加长度为m的区间得到。

状态转移方程：dp[i][j] = max(dp[i - 1][j], dp[i - m][j - 1] + b[i] - b[i - m])

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 5005;
typedef long long ll;
ll dp[MAXN][MAXN], a[MAXN], b[MAXN];
int n, m, k;
int main(int argc, char const *argv[])
{
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= n; ++i) {
        scanf("%I64d", &a[i]);
        b[i] = b[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] + b[i] - b[i - m]);
    printf("%I64d\n", dp[n][k]);
    return 0;
}