C :

#include<stdio.h>
#include<string.h>
#define N 20
int len, m, num[N];//num【】来存放第几个空位的位置
long long d[N][N];//存放第i个空位和第j个空位之间数字是多少
long long ans, sum;//sum是当前的结果
char a[N];//输入
void cal()
{
	sum = 1;
	int x, y;
	int i;
    for(i = 1; i <= m; i++)   //遍历到最后一个空格之后，设定最后一个空格就是最后一个数字之后
	{
		x = num[i-1];
	   	y = num[i];
		sum *= d[x][y];
	}
}
void dfs(int p, int l)
{
	int i;
	int x, y;
	if(l > m-1)   //多于m-1个符号，return
	{
		cal();
		ans = sum > ans ? sum : ans;
		return;
	}
	else
		for(i = p; i < len; i++)   //空位的可能情况是1 ～ len-1
		{
			num[l] = i;
			dfs(i+1, l+1);
		}
}
int main()
{
	int T;
	int i, j;
	scanf("%d", &T);
	while(T--)
	{
		ans = 0;
		memset(d, 0, sizeof(d));
		scanf("%s%d",a,&m);
		if(m == 1)      //剪枝，如果没有乘号那么输出的原数
		{ puts(a); continue; }
		len = strlen(a);
		num[m] = len;   //最后一个空格就是最后一个数字之后
		for(i = 0; i < len; i++)
			for(j = i + 1; j <= len; j++)
			{
				if(j == i + 1) d[i][j] = a[i] - '0';
				else d[i][j] = d[i][j-1] * 10 + a[j-1] - '0';  //第2个空位对应的字母是a【2-1】
			}
		dfs(1, 1);
		printf("%lld\n", ans);
	}
	return 0;
}

C++ :

#include<iostream>
#include<algorithm>
#include<cstring>
#include<ctime>
#include<cmath>
#include<string>
#include<cstdio>
using namespace std;

typedef unsigned long long ULL;

char s[30]; int len;
ULL ans;

ULL getNum(int i, int j) {
    ULL res = 0;
    for(int k = i; k <= j; ++k) res = res * 10 + s[k] - '0';
    return res;
}

void dfs(int pos, int m, ULL res) {
    //cout<<pos<<"--"<<m<<"--"<<res<<endl;
    if(m == 0) {
        ans = max(ans, res * getNum(pos, len-1));
        return;
    }
    for(int i = pos; i <= len; ++i)
        if(len - i - 1 >= m)
        dfs(i+1, m-1, res * getNum(pos, i));
}

int main() {
    int T, m;
    scanf("%d", &T);
    while(T--) {
        ans = 0;
        scanf("%s %d", s, &m);
        len = strlen(s);
        m--;
        if(m == 0) { cout<<s<<endl; continue; }
        for(int i = 0; i < len; ++i)
            if(len - i - 1>= m)
            dfs(i+1, m-1, getNum(0, i));
        cout<<ans<<endl;
    }
    return 0;
}

Java :

import java.util.Scanner;


public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while (t-- > 0) {
			String s = in.next();// 用字符串的方式输入，方便预处理
			s = '0' + s;
			int m = in.nextInt();
			// 进行预处理，用a[i][j]表示原来第i到j段的数值
			int ok = 1;
			long[][] a = new long[s.length()][s.length()];
			for (int i = 1; i < s.length(); i++) {
				if (s.charAt(i) == '0')
					ok = 0;
				for (int j = i; j < s.length(); j++) {// 是要从i开始
					a[i][j] = a[i][j - 1] * 10 + s.charAt(j) - '0';
				}
			}
			if (ok == 0 && s.length() - 1 == m || s.length() - 1 < m) {
				System.out.println(0);
				continue;
			}
			// f[][]表示这个数前i位分成j段的最大乘积
			long[][] f = new long[s.length()+1][s.length()+1];
			// f[i][j] = f[k][j-1]*a[k+1,i];(0<k<i<s.length(),j<m)
			for (int i = 1; i < s.length(); i++)
				f[i][1] = a[1][i];
			long ans = 0;
//			if (m == 1)
//				ans = f[s.length() - 1][1];
			for (int j = 2; j <= m; j++) {//第j个乘号
				for (int i = j; i < s.length(); i++) {
				//	ans = a[i][i];
					// f[i][j] = a[0][s.length()-1-i];
					// System.out.println("a:"+a[0][s.length()-1-i]);
					for (int k = 1; k < i; k++) {
						f[i][j] = Math.max(f[i][j], f[k][j - 1] * a[k + 1][i]);
						// if(ans<f[i][j])
						// ans = f[i][j];
					}
				}
			}
			System.out.println(f[s.length() - 1][m]);
		}
	}
}