HDU Max Sum Plus Plus

Max Sum Plus Plus

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 22   Accepted Submission(s) : 8

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Problem Description

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 S₁, S₂, S₃, S₄ ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + 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 S₁, S₂, S₃ ... 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
#include<iostream>using namespace std;int a[1000005];__int64 b[1000005];__int64 c[1000005];int max(int a,int b){return a>b?a:b;}int main(){int n,m;int i,j;while(~scanf("%d%d",&n,&m)){for(i=0;i<m;i++){scanf("%d",&a[i]);}for(i=1;i<=n;i++){b[i]=c[i]=-INT_MAX;}for(i=0;i<m;i++){for(j=n;j>=1;j--){b[j]=max(b[j],c[j-1])+a[i];c[j]=max(b[j],c[j]);}}printf("%I64d\n",c[n]);}return 0;}