这道题貌似只有@AKEE 大佬A掉，恭喜！

还有因为c++中支持两个参数数量不同的相同名称的函数调用，所以当时就没改成两个函数，这里表示抱歉。

这道题可直接用指针+hash一下，然后就模拟即可。

代码：

#include<bits/stdc++.h>
using namespace std;
const int Mo=10000000;
struct node
{long long int state,ans;node* next;
}*Hash[Mo+10],*p;
long long max(long long a,long long b,long long c,long long d,long long e)
{return max(a,max(b,max(c,max(d,e))));
}
long long max(long long a,long long b,long long c,long long d)
{return max(a,max(b,max(c,d)));
}
long long f(long long x)
{if(x==0)return 0;p=Hash[x%Mo];while(p!=NULL){if(p->state==x)return p->ans;p=p->next;}long long anss=max(x,f(x/2)+f(x/3)+f(x/8)+f(x/9));p=new node;p->state=x;p->ans=anss;p->next=Hash[x%Mo];Hash[x%Mo]=p;return anss;
}
long long f(long long a,long long b)
{long long anss=max(a+b,f(a/2)+f(a/3)+f(a/8)+f(a/9)+b,f(b/2)+f(b/3)+f(b/8)+f(b/9)+a,f(b/2)+f(b/3)+f(b/8)+f(b/9)+f(a/2)+f(a/3)+f(a/8)+f(a/9));return anss;
}
int main()
{//freopen("function.in","r",stdin);//freopen("function.out","w",stdout);long long int a,b;while(cin>>a>>b) {cout<<f(a,b)<<endl;}return 0;
}

另附AKEE大佬代码：（%%%）

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <unordered_map>
#include <algorithm>using namespace std;typedef long long ll;
const int MAXN=10000005;ll n,m,f[MAXN];
unordered_map<ll,ll> has;
ll solve(ll n)
{if(n<=10000000)return f[n];if(has.count(n))return has[n];return has[n]=max(solve(n/2)+solve(n/3)+solve(n/8)+solve(n/9),n);
}int main()
{#ifndef ONLINE_JUDGEfreopen("code.in","r",stdin);//freopen("code.out","w",stdout);#endiff[0]=0;for(int i=1;i<=10000000;i++)f[i]=max(f[i/2]+f[i/3]+f[i/8]+f[i/9],i*1ll);while(cin>>n>>m)cout<<solve(n)+solve(m)<<endl;return 0;
}