/*
dp[i][k] := 截取并获得原字符串前面长度为i+1的字符串，使用K个乘号将它分成K+1个部分，这K+1个部分的乘积的最大值
初始化：
dp[][] = { 0 }
dp[i][0] = Num(0, i); // 截取并获得原字符串前面长度为(i+1)的字符串（字符串0~i位置），返回其所对应的数值
状态方程：
dp[i][k] = max(dp[i][k], dp[j][k-1]*Num(j+1, i))
0<=j<i
答案：
dp[N-1][K]
*/
1 #define _CRTDBG_MAP_ALLOC
2 #include <stdlib.h>
3 #include <crtdbg.h>
4 #define _CRT_SECURE_NO_WARNINGS
5 #define HOME
6
7 #include <iostream>
8 #include <cstdlib>
9 #include <cstdio>
10 #include <cstddef>
11 #include <iterator>
12 #include <algorithm>
13 #include <string>
14 #include <locale>
15 #include <cmath>
16 #include <vector>
17 #include <cstring>
18 using namespace std;
19 const int INF = 0x3f3f3f3f;
20 const int MaxN = 45;
21 const int MaxK = 10;
22
23
24 int N, K;
25 string numStr;
26 int dp[MaxN][MaxK] = { 0 };
27
28
29 int Num(int ibegin, int iend)
30 {
31 int num = 0;
32 int tenMul = 1;
33 for (int i = iend; i >= ibegin; --i)
34 {
35 num += ((numStr[i] - '0') * tenMul);
36 tenMul *= 10;
37 }
38 return num;
39 }
40
41 void Solve()
42 {
43 for (int i = 0; i < N; ++i)
44 {
45 for (int j = 0; j < i; ++j)
46 {
47 for (int k = 1; k <= K; ++k)
48 {
49 //printf("%d -> %d : %d\n", j+1, i, Num(j + 1, i));
50 dp[i][k] = max(dp[i][k], dp[j][k - 1] * Num(j + 1, i));
51 //printf(" %d -> %d : %d\n", j, k-1, dp[j][k-1]);
52 //printf(" %d -> %d : %d\n", i, k, dp[i][k]);
53 }
54 }
55 }
56 cout << dp[N - 1][K] << endl;
57 }
58
59 int main()
60 {
61 #ifdef HOME
62 freopen("in", "r", stdin);
63 //freopen("out", "w", stdout);
64 #endif
65
66 cin >> N >> K;
67 cin >> numStr;
68 for (int i = 0; i < N; ++i)
69 {
70 dp[i][0] = Num(0, i);
71 }
72 Solve();
73
74 #ifdef HOME
75 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
76 _CrtDumpMemoryLeaks();
77 system("pause");
78 #endif
79 return 0;
80 }
