[ABC246D] 2-variable Function

Problem Statement

Given an integer $N$, find the smallest integer $X$ that satisfies all of the conditions below.

• $X$ is greater than or equal to $N$.
• There is a pair of non-negative integers $(a, b)$ such that $X=a^3+a^2b+ab^2+b^3$.

Constraints

• $N$ is an integer.
• $0 \le N \le 10^{18}$

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Input

Input is given from Standard Input in the following format:

$N$

Output

Print the answer as an integer.

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Sample Input 1

9

Sample Output 1

15

For any integer $X$ such that $9 \le X \le 14$, there is no $(a, b)$ that satisfies the condition in the statement.

For $X=15$, $(a,b)=(2,1)$ satisfies the condition.

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Sample Input 2

0

Sample Output 2

0

$N$ itself may satisfy the condition.

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Sample Input 3

999999999989449206

Sample Output 3

1000000000000000000

Input and output may not fit into a $32$-bit integer type.

\(a^3+a^2b+ab^2+b^3=(a+b)^3-2ab(a+b)\)

我们不妨枚举 \(a+b\) 为多少，\((a+b)^3\) 是 \(\theta(n^{\frac{1}{3}})\) 级别的。可以直接枚举。

我们现在知道了 (a+b)，那么我们需要找到是否有 一组 \((a,b)\) 满足上面那个等式。因为和已经确定了，那么剩下的是一个二次函数，具有单调性。我们可以二分 \(a\)，求出 \(b\)，找到最接近的那个解就行了。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL n,ans=2e18,l,r,md;
int main()
{
	scanf("%lld",&n);
	for(LL i=0;i<=2000005;i++)
	{
		if(i*i*i<n)
			continue;
		l=0,r=i/2;
		while(l<=r)
		{
			md=l+r>>1;
			if(i*i*i-2*md*(i-md)*i>=n)
				l=md+1;
			else
				r=md-1;
		}
		if(i*i*i-2*r*(i-r)*i>=n)
			ans=min(ans,i*i*i-2*r*(i-r)*i);
	}
	printf("%lld",ans);
}