CF467C George and Job （DP）

Codeforces Round #267 (Div. 2)

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, ..., p_n. 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, ..., p_n (0 ≤ p_i ≤ 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

题意：

给出n,m,k，给出由n个数组成的序列，在其中选出k组不重叠的连续m个数，使选择的数的和最大。

题解：

DP。

f[i][j],表示在[1,i]中选了j个区间得到的最大的和。

由于要经常求某连续m个数的和，可以先预处理出所有连续m个数的和，sum[i]表示以i为结尾的连续m个数的和。

最后结果为f[n][k]。

状态转移：

    mf1(f);///全部置为-1
    FOR(i,0,n) f[i][0]=0;
    FOR(i,1,n){
        FOR(j,1,k){
            f[i][j]=f[i-1][j];
            if(i-m>=0 && f[i-m][j-1] != -1) f[i][j]=max(f[i][j],f[i-m][j-1]+sum[i]);
        }
    }

 

全代码：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define usll unsigned ll
#define mz(array) memset(array, 0, sizeof(array))
#define mf1(array) memset(array, -1, sizeof(array))
#define minf(array) memset(array, 0x3f, sizeof(array))
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,x,n) for(i=(x);i<=(n);i++)
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define WN(x) printf("%d\n",x);
#define RE  freopen("D.in","r",stdin)
#define WE  freopen("1biao.out","w",stdout)
#define mp make_pair
#define pb push_back
const double eps=1e-10;
const double pi=acos(-1.0);
int a[5111];
ll sum[5111];
ll f[5111][5111];///f[i][j],到i时选了j个区间
int n,m,k;
int main(){
    int i,j;
    RD3(n,m,k);
    FOR(i,1,n){
        RD(a[i]);
    }
    ll sm=0;
    FOR(i,1,n){
        sm+=a[i];
        if(i>=m){
            sum[i]=sm;
            sm-=a[i-m+1];
        }
    }
    mf1(f);///全部置为-1
    FOR(i,0,n) f[i][0]=0;
    FOR(i,1,n){
        FOR(j,1,k){
            f[i][j]=f[i-1][j];
            if(i-m>=0 && f[i-m][j-1] != -1) f[i][j]=max(f[i][j],f[i-m][j-1]+sum[i]);
        }
    }
    printf("%I64d\n",f[n][k]);
    return 0;
}