贪心算法---codeforce680D Bear and Tower of Cubes

最新推荐文章于 2021-02-18 21:44:10 发布

ke_yi_

最新推荐文章于 2021-02-18 21:44:10 发布

阅读量626

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM相关

本文链接：https://blog.youkuaiyun.com/ke_yi_/article/details/81321147

ACM相关专栏收录该内容

47 篇文章

订阅专栏

在这款趣味数学挑战中，小北极熊Limak利用无限的立方体块建造高塔，目标是在不超过指定体积限制的前提下，使塔包含尽可能多的立方体块。玩家需选择最优的总体积X，使得塔的块数最大化且X不超过给定的最大值m。通过贪心策略和深度优先搜索（DFS），算法寻找构建最大塔的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

D. Bear and Tower of Cubes
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
inputCopy
48
outputCopy
9 42
inputCopy
6
outputCopy
6 6

每一次选择最大的x3和次大的x3计算两个中更优的那个，贪心加dfs。

#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int MAXN=1e5+10;
ll x3[MAXN];

pair<ll,ll>ans(0,0);

void dfs(ll m,ll cnt,ll t)
{
    if(!m){
        ans=max(ans,make_pair(cnt,t));
        return ;
    }
    int q=upper_bound(x3+1,x3+100002,m)-x3;
    q--;
    dfs(m-x3[q],cnt+1,t+x3[q]);
    if(q>1)dfs(x3[q]-1-x3[q-1],cnt+1,t+x3[q-1]);
}
int main()
{
    ll m;
    for(ll i=1;i<=100001;i++)
        x3[i]=i*i*i;
    scanf("%I64d",&m);
    dfs(m,0,0);
    printf("%I64d %I64d\n",ans.first,ans.second);
}