码迷,mamicode.com
首页 > 其他好文 > 详细

Codeforces Round #267 (Div. 2) C. George and Job(DP)补题

时间:2014-10-05 00:24:57      阅读:190      评论:0      收藏:0      [点我收藏+]

标签:style   blog   http   color   io   os   ar   for   sp   

                        Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~

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 ≤ nri - li + 1 = m), 

in such a way that the value of sum bubuko.com,布布扣 is maximal possible. Help George to cope with the task.

Input

The first line contains three integers nm and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integersp1, 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
bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #define LL long long
 4 using namespace std;
 5 const int maxn = 5000 + 100;
 6 LL p[maxn],s[maxn],dp[maxn][maxn];
 7 int main(){
 8     int n,m,k;
 9     cin>>n>>m>>k;
10     for(int i = 1;i <= n;i++){
11         cin>>p[i];s[i] = s[i - 1] + p[i];
12     }
13     for(int i = m;i <= n;i++)
14         for(int j = 1;j <= k;j++)
15                 dp[i][j] = max(dp[i-m][j-1]+s[i]-s[i-m],dp[i-1][j]);
16     cout<<dp[n][k]<<endl;
17     return 0;
18 }
View Code

 

Codeforces Round #267 (Div. 2) C. George and Job(DP)补题

标签:style   blog   http   color   io   os   ar   for   sp   

原文地址:http://www.cnblogs.com/jusonalien/p/4006445.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!