1 const int inf = 0x3f3f3f3f;
 2 //dp[i][j]:= 在选择第i个数的情况下前i个数分成j段的最大值
 3 //dp[i][j] = max(dp[i - 1][j] + a[i], max(dp[x][j - 1] -> dp[x][j - 1]) + a[i]) x < i
 4 const int maxn = 1e6 + 5;
 5 int dp[maxn];
 6 int pre[maxn];
 7 int a[maxn], n, m;
 8 
 9 int solve(){
10     memset(dp, 0, sizeof(dp));
11     memset(pre, 0, sizeof(pre));
12     int tmax = -inf;
13     for(int j = 1; j <= m; j++){
14         tmax = -inf;//记录前i个人本次的最大值 
15         for(int i = j; i <= n; i++){
16             dp[i] = max(dp[i - 1] + a[i], pre[i - 1] + a[i]);//使用的是未更新的值，对应j-1的 
17             pre[i - 1] = tmax;    //再更新 
18             tmax = max(tmax, dp[i]);
19         }
20     }
21     return tmax; //不一定是dp[n]，最后一个可能不用选 
22 }
23 
24 int main(){
25     while(scanf("%d %d", &m, &n) != EOF){
26         for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
27         int ans = solve();
28         printf("%d\n", ans);
29     }
30 }

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

 

Sample Output

 

Author

JGShining（极光炫影）