Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 21540 Accepted Submission(s): 7215

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 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 3

2 6 -1 4 -2 3 -2 3

Sample Output

Hint

Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）

dp[i][j]表示前j个数组成的序列最大i子段和，则状态转移方程为

dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]} i-1=<t<j-1

优化1：本题数据范围太大，所以可以用滚动数组优化

优化2：用last数组记录上一层循环得到的最大值，具体见代码

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        #include 
        
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            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   #include 
                   
                    using namespace std;#define eps 1e-8#define randin srand((unsigned int)time(NULL))#define input freopen("input.txt","r",stdin)#define debug(s) cout << "s = " << s << endl;#define outstars cout << "*************" << endl;const double PI = acos(-1.0);const double e = exp(1.0);const int inf = 0x3f3f3f3f;const int INF = 0x7fffffff;typedef long long ll;const int maxn = 1000010;int num[maxn], dp[maxn], last[maxn];int main() { int n, m; while (~scanf("%d %d", &m, &n)) { memset(dp, 0, sizeof(dp)); memset(last, 0, sizeof(last)); for (int i = 1; i <= n; i ++) scanf("%d", &num[i]); int ma; for (int i = 1; i <= m; i ++) { //分成i段 ma = INT_MIN; for (int j = i; j <= n; j ++) { //前j个数 dp[j] = max(dp[j-1], last[j-1]) + num[j];//dp[j]表示前j个划分成i段最大值，last[j-1]表示前j-1个划分成i-1段最大值 last[j-1] = ma; ma = max(ma, dp[j]); } last[n] = ma; } printf("%d\n", ma); } return 0;}