D – 2-variable Function

题目描述：

给你一个n，问大于等于n的最小的x，满足 $x = a^3 + a^2b+ab^2+b^3$

思路：

n最大是1e18,所以a和b都不会超过1e6,再加上固定一个变量a后$f(x) = a^3 + a^2b+ab^2+b^3$,是一个单调递增的函数，所以我们可以枚举a,再去二分找b

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

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
ll n, m, k;

ll cal(ll a, ll b){
    return a * a * a + a * a * b + a * b * b + b * b * b;
}
ll erfen(ll x){
    ll l = 0ll, r = 1000000ll;
    while (l <= r) {
        ll mid = (l + r) / 2;
        if(cal(x, mid) >= n)r = mid - 1;
        else l = mid + 1;
    }
    return cal(x, l);
}

void work(){
    cin >> n;
    ll ans = 1e18;
    for(ll l = 0; l <= 1000000ll; ++l){
        ans = min(ans, erfen(l));
    }
    cout << ans << endl;
}

int main(){
    io;
    work();
    return 0;
}