codeforces 689C C. Mike and Chocolate Thieves(二分)

题目链接：

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 fixedn, 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

题意：

现在有四个小偷,第一个小偷偷走x,第二个x*k,三个x*k*k,四个x*k*k*k;
现在知道一共有n种偷法,问小偷的背包的最小容量是多少;

思路：

二分最小容量,check看这个数能否有n种偷法;

AC代码：

//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef  long long LL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=2e5+10;
const int maxn=1005;
const double eps=1e-10;

LL m,dp[10*N];

int check(LL x)
{
    LL num=0;
   for(int i=2;i<1000006;i++)
   {
       if(dp[i]<=x)
       {
           num+=x/dp[i];
       }
   }
   if(num>m)return 1;
   else if(num==m)return 2;
   return 0;
}
void Init()
{
    for(int i=2;i<=1000006;i++)
    {
        LL temp=i;
        dp[i]=temp*temp*temp;
    }
}
int main()
{
        Init();
        read(m);
        LL l=1,r=inf;
        while(l<=r)
        {
            LL mid=(l+r)>>1;
            if(!check(mid))l=mid+1;
            else r=mid-1;
        }
        if(check(l)==2)cout<<l<<"
";
        else cout<<"-1"<<"
";
        return 0;
}