Algorithm training the largest formula (DP)

Problem 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

:<-----Original link

Dynamic planning, increasing the number of multiplication sign, dp[i][j] is expressed as the maximum value of J multiplication sign for the number of previous I, the position of the J multiplication sign should be discussed at the time of Dp[i][j], if in the K position, then the dp[i][j at this time] Is the first k-1 number has j-1 the maximum value of multiplication sign multiplied by the number of K to J number of sum, and then the dp[i][j] compared to the size, take the maximum (because the position of K has been obtained is the last multiplication sign case, so dp[k-1][j-1] Already represents the maximum value of the previous k-1 number j-1 multiplication sign, directly multiplied by the sum of the remaining numbers)

That is: dp[i][j] = max (dp[i][j],dp[k-1][j-1]* (sum[i]-sum[k)), according to this in turn, note that the number of multiplication sign is less than the number of numbers, and if long long int

#include <stdio.h>#include<string.h>#include<algorithm>using namespacestd;Long Longdp[ -][ -];//Dp[i][j] Represents the maximum value of the first I element J multiplication SignLong Longsum[ -];intMain () {intN, K; intnum; Memset (DP,0,sizeof(DP)); memset (SUM,0,sizeof(sum)); scanf ("%d%d", &n, &k);  for(inti =1; I <= N; i++) {scanf ("%d", &num); Sum[i]= sum[i-1] +num; dp[i][0] =Sum[i]; }     for(inti =2; I <= N; i++)    {         for(intj =1; J <= I-1&& J <= K; J + +)        {             for(intK =2; K <= N; k++)//location of the multiplication sign{Dp[i][j]= Max (Dp[i][j], dp[k-1][j-1]* (sum[i]-sum[k-1])); }}} printf ("%lld\n", Dp[n][k]); return 0;}

View Code

Only hate myself too much water.

Algorithm training the largest formula (DP)