cf_467C_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

情景就不赘述了，给你n个数p1~pn，要你求出这个数列中k段长为m的区间的元素之和的最大值。

要求区间元素和，当然想求出数列的前缀和来了。

二维dp，dp[i][j]表示在取到第i个元素的时候取j个m长度的区间之和的最大值。

dp[i][j]可由dp[i-1][j]不添加区间得到的,或者由dp[i-m][j-1]添加一个区间得到。

遂有状态转移方程：dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+s[i]-s[i-m])，其中s是前缀和。

然后就是简单题了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
long long  dp[5005][5005];
int  main(){
	int  n,m,k;
	scanf ("%d%d%d",&n,&m,&k);//
	long long  a[5005],s[5005];
	memset(a,0,sizeof(a));
	memset(s,0,sizeof(s));
	memset(dp,0,sizeof(dp));
	for (int  i=1;i<=n;i++){
		scanf ("%I64d",&a[i]);
		s[i]=s[i-1]+a[i];
	}
	for (int  i=m;i<=n;i++){
		for (int  j=1;j<=k;j++){
			dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+s[i]-s[i-m]);
		}
	}
	
	printf ("%I64d\n",dp
[k]);
	return 0;
}

（ps：数据爆int） :(