Codeforces round #267 div2 C George and job -- DP

The number of consecutive small sequences with a length of N is divided into K m-long continuous small sequences.

Solution: Apparently DP.

Definition: DP [I] [J] divides the first I elements into j m ends, and I is the largest sum at the end of J.

So there are: DP [I] [J] = max (DP [I-1] [J], DP [I-m] [J-1] + sum [I]-sum [I-m])

5000*5000 space ..

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#define lll __int64using namespace std;lll dp[5002][5001];lll a[5002],sum[5003];int main(){    int n,m,k;    int i,j;    while(scanf("%d%d%d",&n,&m,&k)!=EOF)    {        sum[0] = 0;        for(i=1;i<=n;i++)        {            scanf("%I64d",&a[i]);            sum[i] = sum[i-1]+a[i];        }        for(i=0;i<=n;i++)            for(j=0;j<=k;j++)                dp[i][j] = 0;        for(i=m;i<=n;i++)        {            dp[i][0] = max(dp[i][0],dp[i-1][0]);            for(j=1;j<=k;j++)            {                dp[i][j] = max(dp[i][j],dp[i-1][j]);                dp[i][j] = max(dp[i][j],dp[i-m][j-1]+sum[i]-sum[i-m]);            }        }        cout<<dp[n][k]<<endl;    }    return 0;}

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Codeforces round #267 div2 C George and job -- DP