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(i₂, 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. ^_^ 

InputEach 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. 
OutputOutput the maximal summation described above in one line. 
Sample Input

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

Sample Output

6
8
N个数中去取M个互不相交的区间段，求最大的全体所要的区间段的和。
二维dp降成一维

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 4e18+10;
const int mod = 1000000007;
const int mx = 1e6; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
#define clr(a) memset(a, 0, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pair
void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
ll n, m;
ll a[mx],d[mx],dp[mx][2],mxx[mx],ans;

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    while (~scanf("%lld%lld",&m,&n)){
        ans = -inf;
        re(i, 0, n + 1)dp[i][0] = dp[i][1] = mxx[i] = 0;
        re(i, 1, n+1)sc(a[i]);
        re(j, 1, m + 1) {
            re(i, j, n + 1) {
                if (i == j)dp[i][j & 1] = dp[i - 1][(j - 1) & 1] + a[i];
                else dp[i][j & 1] = max(dp[i - 1][j & 1] + a[i], mxx[i - 1] + a[i]);
                if (i - 1 == j)mxx[i - 1] = dp[i - 1][j & 1];
                else mxx[i - 1] = max(mxx[i - 2], dp[i-1][j& 1]);
            }
        }
        re(i, m, n + 1)ans = max(ans, dp[i][m & 1]);
        printf("%lld\n", ans);
    }
    return 0;
}