Algorithm training the biggest formula

algorithm Training the biggest formulatime limit: 1.0s memory limit: 256.0MBProblem description is simple, give n numbers, do not change their relative position, in the middle add K multiplication sign and n-k-1 a plus, (parentheses arbitrarily add) to make the final result as large as possible. Because multiplication sign and the plus sign are all N-1, there is a sign between each of the two adjacent numbers. For example:
n=5,k=2,5 numbers are 1, 2, 3, 4, 5, respectively, can be added:
1*2* (3+4+5) =24
1* (2+3) * (4+5) =45
(1*2+3) * (4+5) =45
...... Input format input file a total of two lines, the first behavior two spaces separated by an integer, representing N and K, where (2<=n<=15, 0<=k<=n-1). The second behavior is n a number separated by spaces (each number is between 0 and 9). Output format output file only one row contains an integer representing the maximum required result sample Input 5 2
1 2 3 4 5 Sample Output 120 example description (1+2+3) *4*5=120------------------------------------------------

Ideas:

DP[I][J] indicates The maximum value of J multiplication sign for the number of previous I.

Dp[i][j]=max (Dp[i][j],dp[k-1][j-1]*sum (k,i)); K is Any position in the number of previous I.

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ImportJava.util.Scanner; Public classMain { Public Static voidMain (string[] args) {//TODO auto-generated Method Stub        inta[]=New int[120]; Longsum[]=New Long[120];//Sum[i] Indicates the sum of the number of previous I        Longf[][]=New Long[120] [120];//Dp[i][j] Indicates the maximum value when J multiplication sign is in the first I number        intn,k; sum[0]=0; Scanner SC=NewScanner (system.in); N=Sc.nextint (); K=Sc.nextint ();  for(inti=1;i<=n;i++) {A[i]=Sc.nextint (); Sum[i]=sum[i-1]+A[i]; }         for(inti=1;i<=n;i++) f[i][0]=Sum[i];  for(inti=0;i<=n;i++){             intt = math.min (i-1, K);  for(intj=1;j<=t;j++) {//number of control multiply                 for(intk=2;k<=i;k++){                    LongS=SUM[I]-SUM[K-1];//back I-(k-1) number of and, K to <=iF[i][j]=math.max (f[k-1][j-1]*s, f[i][j]);//J indicates the number of products, f[k-1][j-1] to multiply the last is J multiplication sign, K number}}} System.out.println (F[n][k]); }}

Algorithm training the biggest formula