Codeforces Round #356 (Div. 2) D. Bear and Tower of Cubes dfs

D. Bear and Tower of Cubes

题目连接：

http://www.codeforces.com/contest/680/problem/D

Description

Limak is a little polar bear. He plays by building towers from blocks. Every block is a cube with positive integer length of side. Limak has infinitely many blocks of each side length.

A block with side a has volume a3. A tower consisting of blocks with sides a1, a2, ..., ak has the total volume a13 + a23 + ... + ak3.

Limak is going to build a tower. First, he asks you to tell him a positive integer X — the required total volume of the tower. Then, Limak adds new blocks greedily, one by one. Each time he adds the biggest block such that the total volume doesn't exceed X.

Limak asks you to choose X not greater than m. Also, he wants to maximize the number of blocks in the tower at the end (however, he still behaves greedily). Secondarily, he wants to maximize X.

Can you help Limak? Find the maximum number of blocks his tower can have and the maximum X ≤ m that results this number of blocks.

Input

The only line of the input contains one integer m (1 ≤ m ≤ 1015), meaning that Limak wants you to choose X between 1 and m, inclusive.

Output

Print two integers — the maximum number of blocks in the tower and the maximum required total volume X, resulting in the maximum number of blocks.

Sample Input

48

Sample Output

9 42

Hint

## 题意
有一个人在玩堆积木的游戏，给你一个X，这个人会贪心选择一个最大的数，使得这个数a^3<x，然后堆上去，x-=a^3

然后一直重复这个过程，

现在给你一个m，你需要在[1,m]里面找到最大的x，使得使用的数最多，在使用的数最多的情况下，这个数尽量大

-----------------------------------

## 题解：
直接dfs，每次你有两种决策

要么你就使用当前可用的满足x>=p^3

要么你就不使用它，使得当前剩余值等于p^3-1，因为只有这样你才用不上p

然后dfs去处理就好了

代码

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

map<long long ,pair<int ,long long> >H;
long long getmax(long long x)
{
    if(x==1)return 1;
    long long l=1,r=100005,ans=1;
    while(l<=r)
    {
        long long mid = (l+r)/2LL;
        if(mid*mid*mid<=x)ans=mid,l=mid+1;
        else r=mid-1;
    }
    return ans;
}
pair<int,long long> solve(long long x){
    if(x==0)return make_pair(0,0);
    if(H.count(x))return H[x];
    auto &tmp = H[x];
    long long p = getmax(x);
    tmp=solve(x-p*p*p);
    tmp.first++;
    tmp.second+=p*p*p;
    auto tmp2 = solve(p*p*p-1);
    tmp=max(tmp,tmp2);
    return tmp;
}
void QAQ()
{
    long long m;scanf("%lld",&m);
    pair<int,long long>ans=solve(m);
    cout<<ans.first<<" "<<ans.second<<endl;
}
int main()
{
    QAQ();
}