Codeforces 467C George and Job(动态规划)

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 Input

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：这道题给你三个数，n,m,k；n代表有个数pi，然后要在这里面找到长度为m的k对区间。每个区间的界限都有关系：[l1, r1], [l2, r2], …, [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < … < lk ≤ rk ≤ n; ri - li + 1 = m),。就是说不能交叉了。中间可以间隔，也可以不间隔。我觉得这道题最开始想到搜索，思想是：从最开始第一个数，取m个，第m+1个的时候，要么取，要么不取，如果去取了，那么它后面的m-1个也要取，因为我们是取一个区间。类似与搜索的思想。我们这里写的是递推式。就成了动态规划。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL dp[5005][5005];
LL a[5005],sum[5005];

const LL INF=0x3f3f3f3f3f3f3f3f;
int main()
{
LL n,m,k,ans=0,temp;
scanf("%lld%lld%lld",&n,&m,&k);
for(int i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
sum[i]=sum[i-1]+a[i];
}
for(int i= m;i<=n;i++)
{
temp=sum[i]-sum[i-m];
{
for(int j=1;j<=k;j++)
{
dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+temp);
ans=max(ans,dp[i][k]);
}
}
}
printf("%lld\n",ans);
return 0;
}