Codeforces 467C George and Job

最新推荐文章于 2019-04-21 17:45:38 发布

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乔治面临编程工作挑战，需要从给定整数序列中选择特定区间，使得这些区间的和最大化。通过动态规划和前缀和优化解决该问题。

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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 nintegers 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

题解

dp。用一个前缀和优化即可。

#include<iostream> 
#include<cstdio> 
#include<cstdlib> 
#include<cmath> 
#include<cstring>
#define ll long long
using namespace std; 
int n,m,K;
ll sum[5002],f[5002][5002];
void init()
{
	scanf("%d%d%d",&n,&m,&K);
	int i; ll x;
	for(i=1;i<=n;i++)
	   {scanf("%I64d",&x); sum[i]=sum[i-1]+x;}
}
void dp()
{
	int i,j;
	//ll ans;
	for(i=1;i<=n;i++)
	for(j=1;j<=K;j++)
	   {if(j*m>i) continue;
	    f[i][j]=max(f[i][j],f[i-1][j]);
	    f[i][j]=max(f[i][j],f[i-m][j-1]+sum[i]-sum[i-m]);
	    //ans=max(ans,f[i][j]);
	   }
	printf("%I64d\n",f[n][K]);
}
int main() 
{
	init(); dp();
	return 0;
}