cf467C George and Job

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

我是sb……

一道水水的dp

f[i][j]表示前j个取i段的最大价值，t[i]表示从i-m+1到i的区间和

f[i][j]=max(f[i][j-1],f[i-1][j-m]+t[j])

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#define LL long long
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k;
LL f[5010][5010];
LL a[5010],s[5010],t[5010];
int main()
{
	n=read();m=read();k=read();
	for (int i=1;i<=n;i++)a[i]=read();
	for (int i=1;i<=n;i++)s[i]=s[i-1]+a[i];
	for (int i=m;i<=n;i++)
	  t[i]=s[i]-s[i-m];
	for (int i=1;i<=k;i++)
	{
	  for (int j=m;j<=n;j++)f[i][j]=max(f[i][j-1],f[i-1][j-m]+t[j]);
	}
	printf("%I64d",f[k][n]);
}