Max Sum Plus Plus 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, S₂, 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.

又到了被动态规划虐脑的时候了。。

崩溃T^T

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int INF=0x3f3f3f3f;int a[1123456];int dp[1123456];int mm[1123456];int main(){    int n,m;    while(~scanf("%d%d",&m,&n))    {        int i,j;        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            mm[i]=0;            dp[i]=0;        }        dp[0]=0;        mm[0]=0;        int ma;        for(i=1;i<=m;i++)        {            ma=-INF;            for(j=i;j<=n;j++)            {                dp[j]=max(dp[j-1]+a[j],mm[j-1]+a[j]);//dp[j-1]表示的是以j-1结尾i个子段的和..mm[j-1]表示前j-1个元素i-1个子段的和                mm[j-1]=ma;//保证a[i]是一个独立的子段                ma=max(ma,dp[j]);            }        }        printf("%d\n",ma);    }    return 0;}