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 wereexactly 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
1
8
8
54
10
-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：四个小偷想偷巧克力。偷得方案符合如下规则：
巧克力无数，但小偷们的背包容量固定为n(每个小偷最多偷n块巧克力
第一个小偷随意拿，之后三个小偷按照签一个小偷的倍数拿，保证倍数一定。
譬如 1,2,4,8，第一个拿1块，之后每人拿前一人的二倍。 倍数>1
给出方案数目，问背包容量n为多少可以偷出这么多方案。
如果有多解，输出最少的背包容量。
分析：
先二分背包容量，
再枚举倍数的话，其实对于已知容量n，已知倍数x下。
能偷得方案数其实就是n/x 自然取整即可（求出此时的方案数）
方案数跟m比较 ，然后继续二分。
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
long long L=,R=1e16;
long long check(long long m)
{
long long sum=;
for(long long i=;;i++)
{
if(m>=i*i*i)
sum+=m/(i*i*i);
}
return sum;
}
int main()
{
long long n,mid,s,ans=-;;
scanf("%lld",&n);
while(L<=R)
{
mid=(L+R)/;
s=check(mid);
if(s==n) ans=mid;
else if(s>=n) R=mid-;
else L=mid+;
}
cout<<ans<<endl;
return ;
}