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,,n) f[i][]=;
     FOR(i,,n){
         FOR(j,,k){
             f[i][j]=f[i-][j];
             if(i-m>= && f[i-m][j-] != -) f[i][j]=max(f[i][j],f[i-m][j-]+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-;
 const double pi=acos(-1.0);
 int a[];
 ll sum[];
 ll f[][];///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,,n){
         RD(a[i]);
     }
     ll sm=;
     FOR(i,,n){
         sm+=a[i];
         if(i>=m){
             sum[i]=sm;
             sm-=a[i-m+];
         }
     }
     mf1(f);///全部置为-1
     FOR(i,,n) f[i][]=;
     FOR(i,,n){
         FOR(j,,k){
             f[i][j]=f[i-][j];
             if(i-m>= && f[i-m][j-] != -) f[i][j]=max(f[i][j],f[i-m][j-]+sum[i]);
         }
     }
     printf("%I64d\n",f[n][k]);
     return ;
 }