https://codeforces.com/contest/1143/problem/B

题解：数位DP 

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int shu[20], dp[20];

int dfs(int len, bool shangxian)
{
    if (len == 0)
        return 1;
    if (!shangxian && dp[len])   //为什么要返回呢？可以画图理解当我们搜到3XXX时，程序运行到1XXX时就已经把3XXX之后的搜索完了，记忆化也是这个用意.
        return dp[len];
    int res= 0, maxx = (shangxian ? shu[len] : 9);
    for (int i = 0; i <= maxx; i++)
    {
        if(i==0)
            res=max(res,dfs(len - 1,shangxian && i == maxx));  //只有之前有限制现在的达到了上限才能构成限制
        else
            res=max(res,i*dfs(len - 1,shangxian && i == maxx));
    }
    return shangxian ? res: dp[len] = res; //如果有限制，那么就不能记忆化，否则记忆的是个错误的数.
}

int solve(int x)
{
    memset(shu, 0, sizeof(shu));
    int k = 0;
    while (x)
    {
        shu[++k] = x % 10;  //保存a,b的数
        x /= 10;
    }
    return dfs(k, true);
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    cout<<solve(n)<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}