Codeforces Round #361 (Div. 2) C 二分

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C. Mike and Chocolate Thieves

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

题意：

给出一个数m，一个数x以k(k>1)倍增加三次形成四个数，所有的x及k都需要枚举到

问正好有m组这样的四元组所对应的最大的第四个数是多少

思路：

其实就是二分答案，然后判断这个数符合不符合条件

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
ll m;
ll a[200010];
ll solve(ll x){ll ans=0;for(int i=1;i<=200000;i++){ans+=x/a[i];if(ans>m) break;}return ans;
}
int main(){for(int i=1;i<=200000;i++) a[i]=(ll)(i+1)*(i+1)*(i+1);scanf("%I64d",&m);ll ans=-1,l=1,r=1e16,mid;while(l<r){mid=(l+r)/2;if(solve(mid)==m) {ans=mid;r=mid;}else if(solve(mid)<m){l=mid+1;}else r=mid;}printf("%I64d\n",ans);return 0;
}

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