CodeForces - 467C


George and Job



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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



[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 11 2 3 4 5




Output


9




Input


7 1 32 10 7 18 5 33 0




Output


61


 



#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 100010
#define ll long long
using namespace std;
ll a[N];
ll dp[2][N];
ll sum[N];
int main()
{
	int n,m,k;
	while(scanf("%d%d%d",&n,&m,&k)!=EOF)
	{
		ll ans=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			sum[i]=0;
			ans+=a[i];
			if(i<m)
				continue;
			for(int j=i;j>i-m;j--)
				sum[i]+=a[j];
		}
		if(n==k||n==m)
		{
			printf("%lld\n",ans);
			continue;
		}
		memset(dp,0,sizeof(dp));
		ans=0;
		for(int l=1;l<=k;l++)
		{
			for(int i=l*m;i<=n;i++)
			{
				dp[l&1][i]=max(dp[l&1][i],dp[l&1][i-1]);
				dp[l&1][i]=max(dp[l&1][i],dp[(l-1)&1][i-m]+sum[i]);
				ans=max(ans,dp[l&1][i]);
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
}






Time Limit: 1000MS

 

Memory Limit: 262144KB

 

64bit IO Format: %I64d & %I64u