Codeforces round #267 (Div. 2) C. George and job (DP) Supplement

Codeforces round #267 (Div. 2) C. George and job question link please click ~

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 sequenceNIntegersP_{1, bytes,P2, middle..., middle ,...,PN. You are to chooseKPairs of integers:}

 

[ L _{1, bytes,R1], region [L2, bytes,R2], numbers..., numbers [LK, Bytes,RK] (1 limit ≤ limitL1 bytes ≤ bytesR1 worker <workerL2 bytes ≤ bytesR2 rows <values... rows <valuesLKLimit ≤ limitRKLimit ≤ limitN;RIAccept-Encoding-LIBytes + bytes 1 records = bytesM), Then ),}

In such a way that the value of sum is maximal possible. Help George to interact with the task.

Input

The first line contains three IntegersN,MAndK(1 limit ≤ limit (MLimit × limitK) Bytes ≤ bytesNLimit ≤00005000). The second line containsNIntegersP_{1, bytes,P2, middle..., middle ,...,PN(0 bytes ≤ bytesPILimit ≤ limit 10^{9 ).}}

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

 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 }

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Codeforces round #267 (Div. 2) C. George and job (DP) Supplement