The largest product range DP--

Memory limit: 128 MiB time limit: 1000 ms standard output

Question Types: traditional evaluation methods: text comparison

This year is the International Mathematical Union to determine the "2000-- World Mathematics", also coincides with the 90th anniversary of a famous mathematician, Mr. Hua birthday. Mr. Hua's hometown Jintan, math quiz organized activity of a spectacular, one of your good friend XZ also had the honor to participate. Activity, the host to all the players participating in a such a Title: with a length of N strings of numbers, the player requires the use of the K multiplication sign it into K + 1 parts, to find a points, K + 1 such that the product of these portions is maximized. Meanwhile, in order to help the player can correctly understanding the problem, a moderator is also cited in the following examples: a string of numbers: 312, when N = 3, K = 1 there will be the following two points Method: 1) 3 12 = 362) 31 2 = 62 in this case, the result is in line with requirements of the subject: 31 * 2 = 62. Now, you help your friends XZ design a program to obtain the correct answer.

There are a first line of two natural numbers N, K (6≤N≤40,1≤K≤6)

The second line is a numeric string of length N.

The product obtained maximum output (a natural number).

Sample input

4 2
1231

Sample Output

. 1 #include <the iostream>
 2 #include <CString>
 . 3 #include <cstdio>
 . 4 #include <Stack>
 . 5  the using  namespace STD;
 . 6 unsigned Long  Long n-, / * K th multiplication, k + 1 partial * / DP [ 50 ] [ 10 ], A [ 41 is ] [ 41 is ] / * before i j bits into segments * / , Ink;
 . 7  int K6;
 . 8  char  in [ 51 is ];
 . 9  int main ()
 10  {
 . 11     cin>>n>>k6;
12     k6++;
13     cin>>in+1;
14     for(int i=1;i<=n;i++)
15     {
16         ink=0;
17         for(int j=i;j<=n;j++)
18         {
19             ink=ink*10+in[j]-'0';
20             A[i][j]=ink;
21         }
22     }
23      dp[0][0]=1;
24     for(int i=1;i<=n;i++)/*dp[i][j]-->前i位分成j段*/
25     {
26         int ki=min(i,k6);
27         for(int j=1;j<=ki;j++)
28         {
29             for(int k=1;k<=i;k++)
30             {
31                 if(dp[k-1][j-1]*A[k][i]>dp[i][j])
32                 {
33                     dp[i][j]=dp[k-1][j-1]*A[k][i];
34                 }
35             }
36         }
37     }
38     cout<<dp[n][k6];
39 }