Codeforces 689C Mike and Chocolate Thieves【二分+思维】

C. Mike and Chocolate Thieves

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

Input

1

Output

8

Input

8

Output

54

Input

10

Output

-1

Note
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.

题目大意：

给你一个数N，表示要找到N个四元组，其中（a1，a2，a3，a4）是一个等比数列。对应N个四元组中，那个最大的数记做maxn，

我们的任务就是要寻找最小的这个maxn；

思路：

1、假设我们枚举这个maxn，我们肯定是随着maxn的增大，四元组的个数也就随着增多，那么我们考虑到这里的单调性，我们可以进行二分查找的方法来枚举这个maxn。

2、对应当前枚举出来的二分值mid，我们只要统计小于等于mid的四元组一共存在多少个即可。

考虑这样一点，a4一定是一个K*L^3的形式存在的，那么我们要找的存在的个数，也就是ΣK*L.

我们可以预处理出10^6以内的所有数字的三次方是多大，并且已知最大是10^18，接近于LL的极限。

那么我们可以通过枚举L再通过除法来计算当前情况下的K。不断统计即可。

统计过程中可以加入一些小剪枝，参考代码即可，不做啰嗦。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
ll three[1000005];
ll n,ans;
int Slove(ll mid)
{
ll cont=0;
for(int i=2;i<=10000000;i++)
{
if(three[i]>mid)break;
else
{
cont+=mid/three[i];
}
if(cont>n)return 1;
}
if(cont==n)
{
ans=mid;
return 1;
}
else return 0;
}
void init()
{
for(ll i=2;i<=1000000;i++)
{
three[i]=i*i*i;
}
}
int main()
{
init();
while(~scanf("%I64d",&n))
{
ans=-1;
ll l=1;
ll r=1000000000000000000;
while(r-l>=0)
{
ll mid=(l+r)/2;
if(Slove(mid)==1)
{
r=mid-1;
}
else l=mid+1;
}
printf("%I64d\n",ans);
}
}