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()
{
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_JUDGE
freopen("code.in","r",stdin);
#endif
f
[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;
}
转载于:https://www.cnblogs.com/vercont/p/10210031.html
相关资源：JAVA上百实例源码以及开源项目
