upc组队赛12 Cardboard Container【枚举】

Cardboard Container

Problem Description

fidget spinners are so 2017; this years’ rage are fidget cubes. A fidget cube is a cube with unit side lengths, which you hold in your hand and fidget with. Kids these days,right?

You work in the planning department for a company that creates and ships fidget cubes. Having done some market analysis, you found that your customers want to receive shipments of exactly V fidget cubes.

This means you have to design a container that will hold exactly V fidget cubes. Since fidget cubes are very fragile, you cannot have any empty space in your container. If there is empty space, they might move around, bump into each other and get damaged. Because of this, you decide to ship the fidget cubes in a rectangular cardboard box.

The cost of a cardboard box is proportional to its surface area, costing exactly one unit of money per square unit of surface area. Of course you want to spend as little money as possible. Subject to the above constraints, how much money do you have to spend on a box for V fidget cubes?

Input

Input

The input contains a single integer, 1 ≤ V ≤ 106, the number of fidget cubes for which you need to build a box.

Output

Print the cost of the cheapest rectangular box as specified in the statement.

Sample Input

1

Sample Output

6

题意

求n个立方体小块叠成一个大立方体形成的最小表面积

题解

枚举长和宽，表面积 = 2*（（长+宽）+（长+高）+（宽+高）），取最小的

代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define scac(x) scanf("%c",&x)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define pri3(x,y,z) printf("%d %d %d\n",x,y,z)
#define prl(x) printf("%lld\n",x)
#define prl2(x,y) printf("%lld %lld\n",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld\n",x,y,z)
#define ll long long
#define LL long long
inline ll read(){ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;}
#define read read()
#define pb push_back
#define mp make_pair
#define P pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define MOD 998244353
#define mod 1e9+7
#define N 2000005
const int maxn=2e5+5;
int n;
ll ans;
int main()
{
	n = read;
	ans = INF;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n/i && j <= i; j++){
			if(n%(i * j)!=0) continue;
			int k = n/(i * j);
			ans = min(ans,(ll)2*(i*j + i*k + j*k));
		}
	prl(ans);
}