Codeforces Round #361 (Div. 2) C. Mike and Chocolate Thieves

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.

二分答案，然后主要是计算这个有多少种方式。
首先是求乘以2的
1，2，4，8
2，4，8，16
3，6，12，24
4，8，16，32
诸如此类，我们可以发现，只要最后一个数不会大于二分的数字，就是这个乘积的立方乘以首个数字，就可以用 tans+=(val/j);求得。

代码写的好丑啊！

#include<bits/stdc++.h>using namespace std;typedef __int64 ll;const ll maxll=(~((unsigned long long)0))>>1;ll n;ll f(ll val) {    ll tans=0;    for(ll i=2; ; ++i) {        ll j=i*i*i;        if(j>0&&j<=val) {            tans+=(val/j);            if(tans>n)return n+1;        } else break;    }    return tans;}ll work() {    ll left=8,right=maxll,mid,best=maxll;    while(left<=right) {        mid=left+((right-left)>>1);        ll t=f(mid);        if(t==n) best=min(best,mid),right=mid-1;        else if(t>n)right=mid-1;        else left=mid+1;    }    return best==maxll?-1LL:best;}int main() {    cin>>n;    ll ans=work();    if(ans==-1LL)return 0*printf("-1\n");    for(;; --ans) {        if(f(ans)<n) {            return 0*printf("%I64d\n",ans+1);        }    }    return 0;}