题目大意：从n个数中选出m段不相交的子串，子串的长度均为k，问所有选出来的子串的所有数的和最大为多少。

DP题，DP还是太弱，开始时的dp方程居然写成了O(n^3)...  

dp[i][j]:  以num[i]结尾的序列，分成j段的最大和

dp[i][j]=max(dp[k][j-1]+sum[i]-sum[i-m])  这样的话，其实只要第一重循环是选的段数，第二重循环时数字个数

我又换了种思路

dp[i][j]: 前i个数，分成j段的最大和

dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+sum[i]-sum[i-m])

思路：二维嘛，所以写出来的dp方程肯定是要么从第一维变化转移过来的，要么从第二维的变化转移过来的

AC代码：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1);
const int INF = 100000000;
const int MAXN = 5000+100;

ll num[MAXN],dp[MAXN][MAXN],pp[MAXN];
int n,m,k;

ll solve()
{
    CL(dp,0);
    for(int i=m;i<=n;i++)
        for(int j=1;j<=k;j++)
            dp[i][j]=max(dp[i-m][j-1]+pp[i]-pp[i-m],dp[i-1][j]);
    return dp[n][k];
}

int main()
{
    //IN("C.txt");
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        pp[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&num[i]);
            pp[i]=pp[i-1]+num[i];
        }
        printf("%I64d\n",solve());
    }
    return 0;
}

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<string>  
#include<algorithm>  
#include<map>  
#include<queue>  
#include<stack>  
#include<set>  
#include<cmath>  
#include<vector>  
#define inf 0x3f3f3f3f  
#define Inf 0x3FFFFFFFFFFFFFFFLL  
#define eps 1e-8  
#define pi acos(-1.0)  
using namespace std;  
typedef long long ll;  
const int maxn = 5000 + 5;  
int a[maxn];  
ll sum[maxn],dp[maxn][maxn],maxv[maxn];  
int main()  
{  
//    freopen("in.txt","r",stdin);  
//    freopen("out.txt","w",stdout);  
    int n,m,k;  
    scanf("%d%d%d",&n,&m,&k);  
    for(int i = 1;i <= n;++i)  
        scanf("%d",&a[i]);  
    sum[0] = 0;  
    for(int i = 1;i <= n;++i)  
        sum[i] = sum[i-1] + a[i];  
    memset(dp,0xff,sizeof(dp));  
    memset(maxv,0,sizeof(maxv));  
    dp[0][0] = 0;  
    for(int j = 1;j <= k;++j)  
    {  
        for(int i = 1;i <= n;++i)  
        {  
            if(i - m >= 0)  
            {  
                dp[i][j] = max(dp[i][j],maxv[i-m] + sum[i] - sum[i-m]);  
            }  
        }  
        for(int i = 1;i <= n;++i)  
            maxv[i] = max(maxv[i-1],dp[i][j]);  
    }  
    ll ans = 0;  
    for(int i = 1;i <= n;++i)  
        if(dp[i][k] != -1)  
            ans = max(ans,dp[i][k]);  
    printf("%I64d\n",ans);  
    return 0;  
}  

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