CF467C： George and Job（dp）

最新推荐文章于 2020-05-27 22:21:56 发布

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解决一个程序员面临的挑战：从给定的整数序列中找出k个不重叠且长度为m的连续区间，使得这些区间的元素之和达到最大值。通过动态规划的方法实现了这一目标。

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 $\text{[math]}$ 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.

Examples

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：给定N个数，要求找出k个不重叠且长度为m的连续区间，使他们的和最大。

思路：dp[i][j]表示到i位置时找出j组所得的最大值。

# include <stdio.h>
# include <string.h>
# include <algorithm>
using namespace std;
long long a[5001]={0}, dp[5001][5001];
int main()
{
    int n, m, k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        memset(dp, 0, sizeof(dp));
        for(int i=1; i<=n; ++i)
        {
            scanf("%d",&a[i]);
            a[i] += a[i-1];
        }
        for(int i=1; i<=k; ++i)
            for(int j=i*m; j<=n; ++j)
                dp[j][i] = max(dp[j-1][i], dp[j-m][i-1] + a[j] - a[j-m]);
        printf("%I64d\n",dp[n][k]);
    }
    return 0;
}