Codeforce 467C. George and Job(DP)_c. george and number codeforce-优快云博客

本文介绍了一道经典的编程挑战题，题目要求从一个整数序列中选择k组连续子序列，每组长度为m，使得这k组子序列的元素之和最大。文章提供了完整的AC代码实现，采用动态规划算法解决此问题。

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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.

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

题目大意：一个长度为n的序列选k组，每组长度为m，问最优的选法

ac代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#define LL long long
using namespace std;
LL sum[5050];
LL dp[5050][5050];
int n,m,k;
LL DP()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    for(i=m;i<=n;i++)
    {
        for(j=1;j<=i&&j<=k;j++)
        {
            dp[i][j]=max(dp[i-m][j-1]+(sum[i]-sum[i-m]),dp[i-1][j]);
        }
    }
    return dp[n][k];
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        int i;
        memset(sum,0,sizeof(sum));
        for(i=1;i<=n;i++)
        {
            LL x;
            scanf("%lld",&x);
            sum[i]=sum[i-1]+x;
        }
        printf("%lld\n",DP());
    }
}