Codeforces Round #267 (Div. 2) C

標籤：演算法   acm   codeforces   algorithm   hdu   

題目：C. George and Jobtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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],?...,?[ l k,? r k] (1?≤? l1?≤? r1?<? l2?≤? r2?<?...?<? l k?≤? r k?≤? n;  r i?-? l i?+?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

題意分析：DP一下。dp[I][j]存前i有j組的最大值。
狀態轉移方程：

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項a[i]的和。
代碼:

#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 ;}

Codeforces Round #267 (Div. 2) C