HDU

 Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (1 ≤ i ≤ j ≤ n).Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x is not allowed).But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^

Input
Each test case will begin with two integers m and n, followed by n integers S 1, S 2, S 3 … S n.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input

1 3 1 2 32 6 -1 4 -2 3 -2 3

Sample Output

68

Hint

Huge input, scanf and dynamic programming is recommended.

题解参考http://blog.sina.com.cn/s/blog_677a3eb30100jxqa.html

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <queue>#define N 1000100#define INF 0x3f3f3f3fusing namespace std;long long dp[2][N];long long a[N];int main(){    int n, m;    while(scanf("%d%d", &m, &n) != EOF)    {        memset(dp, 0, sizeof(dp));        for(int i = 1; i <= n; i++)        {            scanf("%lld", &a[i]);        }        for(int i = 1; i <= m; i++)        {            dp[i % 2][i] = dp[(i + 1)% 2][i- 1] + a[i];            long long int _max = dp[(i + 1) % 2][i - 1];            for(int j = i + 1; j <= n; j++)            {                _max = max(_max, dp[(i + 1) % 2][j - 1]);                dp[i % 2][j] = max(dp[i % 2][j - 1], _max) + a[j];            }        }        long long int _max = -INF;        int index = m % 2;        for(int i = m; i <= n; i++)        {            _max = max(_max, dp[index][i]);        }        printf("%lld\n", _max);    }    return 0;}