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

                            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  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
     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
    额 继续努力吧 骚年~~~
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  • 原文地址:https://www.cnblogs.com/jusonalien/p/4006445.html
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