CodeForces - 689C Mike and Chocolate Thieves (二分)

针对四名盗贼按特定比例分配巧克力的问题,本篇介绍了一种通过二分查找结合枚举的方法来确定盗贼们携带巧克力的最大数量。该文详细解释了如何在限定条件下找出符合条件的最小n值。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目链接:http://codeforces.com/problemset/problem/689/C点击打开链接

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 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.


求恰好m个满足式子x*k*k*k的最小值

二分个数然后枚举k 计算个数

因为是log级别因此可以到1e15

#include <bits/stdc++.h>
using namespace std;
long long int m;
long long int cal(long long int num)
{
    long long int ans=0;
    for(long long int i=2;i*i*i<=num;i++)
        ans+=(num/(i*i*i));
    return ans;
}
int  main()
{
    long long int l=0;
    long long int r=1e16;
    long long int ans=0;
    cin >> m;
    while(l<=r)
    {
        long long int mid=l+(r-l)/2;
        if(cal(mid)>=m)
        {
            ans=mid;
            r=mid-1;
        }
        else
        {
            l=mid+1;
        }
    }
    if(cal(ans)==m)
        printf("%lld",ans);
    else
        printf("-1");
}



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值