Codeforces Round #267 (Div. 2) C

题目：

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 11 2 3 4 5

output

9

input

7 1 32 10 7 18 5 33 0

output

61

题意分析：

DP一下。dp[I][j]存前i有j组的最大值。

状态转移方程：

if(i-m>=0)dp[i][j] = max(dp[i][j],max(dp[i-1][j],dp[i-m][j-1]+b[i]-b[i-m])) ;

else 

<span style="white-space:pre"></span>dp[i][j] = max(dp[i][j],dp[i-1][j]) ;

其中b[i]存前i项a[i]的和。

代码:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;long long dp[5005][5005] ;long long a[65555] ;long long b[65555] ;int main(){int n,m,k ;cin>>n>>m>>k;int i,j;for(i=1;i<=n;i++){cin>>a[i];b[i] = b[i-1] + a[i] ;}for(i=1;i<=n;i++){for(j=1;j<=k;j++){if(i-m>=0)dp[i][j] = max(dp[i][j],max(dp[i-1][j],dp[i-m][j-1]+b[i]-b[i-m])) ;else dp[i][j] = max(dp[i][j],dp[i-1][j]) ;}}cout<<dp[n][k]<<endl;return 0 ;}