高精度乘法+DP

令$dp[i][j]$表示对于前$i$个字符，插入$j$个乘号所能获得的最大值，显然区间DP

#include<bits/stdc++.h>
using namespace std;

int n,k;
vector<int> a;
vector<vector<vector<int> > > dp(45,vector<vector<int> >(10)),num(45,vector<vector<int> >(45));
vector<int> operator * (vector<int> A,vector<int> B) //I hate high precision 我恨高精度
{
    vector<int> C(A.size()+B.size(),0);
    for (int i=0;i<A.size();i++)
    {
	    for (int j=0;j<B.size();j++)
        {
		    C[i+j]+=A[i]*B[j];
		    C[i+j+1]+=C[i+j]/10;
		    C[i+j]%=10;
		}
    }
    while (C.size()>1 && C.back()==0) C.pop_back();
    return C;
}
bool operator < (vector<int> A,vector<int> B)
{
    if (A.size()!=B.size()) return A.size()<B.size();
    for (int i=A.size()-1;i>=0;i--)
    {
        if (A[i]!=B[i])
            return A[i]<B[i];
    }
    return 0;
}
vector<int> max(vector<int> A,vector<int> B)
{
    return (A<B ? B:A);
}
int main()
{
    string s;
	cin >> n >> k >> s;
    for (int i=s.size()-1;i>=0;i--)
    	a.push_back(s[i]-'0');
    for (int i=0;i<n;i++)
    {
        num[i][i].push_back(a[i]);
        for (int j=i+1;j<n;j++)
            num[i][j]=num[i][j-1],num[i][j].push_back(a[j]);
    }
    for (int i=0;i<n;i++)
        dp[i][0]=num[0][i];
    for (int i=0;i<n;i++)
    {
        for (int j=1;j<=k;j++)
        {
            for (int l=i;l>0;l--)
                dp[i][j]=max(dp[i][j],dp[l-1][j-1]*num[l][i]);
        }
    }
    for (auto it=dp[n-1][k].end()-1;it>=dp[n-1][k].begin();it--)
        cout << *it;
	return 0;
}