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.
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 題意：
給出一列數（0<=p<=10^9) 求出k 個區間間長為m 的最大總和，且選擇的區間不能有交集。
思路：
類似于背包問題。把連續的m個數看成一個物品，以取的區間個數作為狀态。本題要解決的重要問題是如何是得到的區間沒有交集——使用字首和數組！
dp[i][j] = max ( dp[i-1][j], dp[i-1][j-1] + s[i] - s[i-m] )
表示前i 個數 j 個區間的最大和。答案為 dp[n][k] 。
CODE：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
const int inf=0xfffffff;
typedef long long ll;
using namespace std;
ll p[5005], s[5005];
ll dp[5005][5005];
int main()
{
//freopen("in","r",stdin);
int n, m, k;
s[0] = 0;
while(~scanf("%d %d %d", &n, &m, &k)){
for(int i = 1; i <= n; i ++){
scanf("%I64d", &p[i]);
s[i] = s[i-1] + p[i];
}
memset(dp, 0, sizeof(dp));
for(int i = m; i <= n; i ++){
for(int j = 1; j <= k; j ++){
dp[i][j] = max(dp[i - 1][j], dp[i-m][j-1] + s[i] - s[i - m]);
}
}
printf("%I64d\n", dp[n][k]);
}
return 0;
}