CodeForces 689C Mike and Chocolate Thieves （二分+数论）

Mike and Chocolate Thieves

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121333#problem/G

Description

Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible!

Aside from loving sweet things, thieves from this area are known to be very greedy. So after a thief takes his number of chocolates for himself, the next thief will take exactly k times more than the previous one. The value of k (k > 1) is a secret integer known only to them. It is also known that each thief's bag can carry at most n chocolates (if they intend to take more, the deal is cancelled) and that there were exactly four thieves involved.

Sadly, only the thieves know the value of n, but rumours say that the numbers of ways they could have taken the chocolates (for a fixed n, but not fixed k) is m. Two ways are considered different if one of the thieves (they should be numbered in the order they take chocolates) took different number of chocolates in them.

Mike want to track the thieves down, so he wants to know what their bags are and value of n will help him in that. Please find the smallest possible value of n or tell him that the rumors are false and there is no such n.

Input

The single line of input contains the integer m(1 ≤ m ≤ 1015) — the number of ways the thieves might steal the chocolates, as rumours say.

Output

Print the only integer n — the maximum amount of chocolates that thieves' bags can carry. If there are more than one n satisfying the rumors, print the smallest one.

If there is no such n for a false-rumoured m, print  - 1.

Sample Input

Input

1

Output

8

Input

8

Output

54

Input

10

Output

-1

Hint

In the first sample case the smallest n that leads to exactly one way of stealing chocolates is n = 8, whereas the amounts of stealed chocolates are (1, 2, 4, 8) (the number of chocolates stolen by each of the thieves).

In the second sample case the smallest n that leads to exactly 8 ways is n = 54 with the possibilities: (1, 2, 4, 8),  (1, 3, 9, 27),  (2, 4, 8, 16),  (2, 6, 18, 54),  (3, 6, 12, 24),  (4, 8, 16, 32),  (5, 10, 20, 40),  (6, 12, 24, 48).

There is no n leading to exactly 10 ways of stealing chocolates in the third sample case.

题意：

四个数满足 a, b=ak, c=akk, d=akkk <= n 且a>0 k>1；

已知满足要求的a k组合的方案数为m，求最小的n；

题解：

二分答案n：(范围1-1e18)

根据akk*k<=n的条件枚举k(枚举a规模太大)，并对可行的a计数；

将方案数与m比较；

注意：

对于同一个m可能存在多个满足条件的n，由于要求最小的n，

则当找到一个n使得条件满足时，应继续向左区间递归找下一个更小的n，而不是停止检索.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 1100
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int main(int argc, char const *argv[])
{
    //IN;

    LL m;
    while(scanf("%I64d",&m) != EOF)
    {
        LL ans, minans = -1;
        LL l=1, r=(1LL<<60);
        while(l <= r) {
            LL mid = (l+r) / 2;
            ans = 0;
            for(LL k=2; k*k*k<=mid; k++) {
                ans += mid / (k*k*k);
            }

            if(ans==m) minans = mid;
            if(ans >= m) r = mid - 1;
            else l = mid + 1;
        }

        printf("%I64d\n", minans);
    }

    return 0;
}