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?≤?n; ri?-?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 n, m and k (1?≤?(m?×?k)?≤?n?≤?5000). The second line contains n integers p1,?p2,?…,?pn (0?≤?pi?≤?109).

Output
Print an integer in a single line — the maximum possible value of sum.

Examples
Input
5 2 1
1 2 3 4 5
Output
9
Input
7 1 3
2 10 7 18 5 33 0
Output
61
題意：從n個數(shù)中選出k段長度為m的連續(xù)數(shù)組，使得這k段加和最大。
所選取的段數(shù)可以是連續(xù)的，也可以不是連續(xù)的。我們使dp[i][j]代表著從前i個數(shù)字中選取j段的最大值。對于dp[i][j]來說，它可以是前i-1個數(shù)選取j段，也可以是前i-m段選取j-1段之后，再選取一段。
狀態(tài)轉(zhuǎn)移方程：dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+cnt);//cnt為從i-m到i的數(shù)字和。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=5e3+100; ll a[maxx],sum[maxx]; ll dp[maxx][maxx]; int n,m,k;int main() {scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=n;i++) scanf("%lld",&a[i]),sum[i]=sum[i-1]+a[i];int x=m-1;ll cnt;for(int i=m;i<=n;i++){cnt=sum[i]-sum[i-x-1];for(int j=1;j<=k;j++){dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+cnt);}}printf("%lld\n",dp[n][k]); }

努力加油a啊，(^o)/~

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