hdu_4430_策略二分经典

最新推荐文章于 2018-04-14 12:41:47 发布

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本文介绍了一种算法，用于解决在给定总数的情况下，如何将蜡烛排列成多个同心圆的问题。通过限定条件和使用二分查找技术，找到使蛋糕看起来更美观的最优蜡烛排列方案。

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Today is Yukari’s n-th birthday. Ran and Chen hold a celebration party for her. Now comes the most important part, birthday cake! But it’s a big challenge for them to place n candles on the top of the cake. As Yukari has lived for such a long long time, though she herself insists that she is a 17-year-old girl.
To make the birthday cake look more beautiful, Ran and Chen decide to place them like r ≥ 1 concentric circles. They place ki candles equidistantly on the i-th circle, where k ≥ 2, 1 ≤ i ≤ r. And it’s optional to place at most one candle at the center of the cake. In case that there are a lot of different pairs of r and k satisfying these restrictions, they want to minimize r × k. If there is still a tie, minimize r.

Input
There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 1012.

Output
For each test case, output r and k.

Sample Input
18
111
1111

Sample Output
1 17
2 10
3 10
题意:
给蛋糕插r圈蜡烛,中间插不插都可以,第i圈插k的i次方个,然后求在r*k(值相同时取r最小)最小的情况下的r ,k值已知总蜡烛数n
解:
由题得蜡烛的总数可近似为一个等差数列,知道r最大为40多.
然后对每个r二分求解k,注意二分时k的范围;

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll solve(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans*=a;
        b>>=1;
        a*=a;
    }
    return ans;
}
int main()
{
    ll n,i,j;
    while(cin>>n)
    {
        ll a=1,b=n-1;
        for(i=2;i<41;i++)
        {
            ll low=2,high=pow(n,1.0/i),k;
            while(low<=high)
            {
                k=(low+high)/2;
                ll ans=k*(1-solve(k,i))/(1-k);
                if(ans==n||ans==n-1)
                {
                    if(k*i<a*b)
                    {
                        a=i;b=k;
                    }
                }
                if(ans>n)
                    high=k-1;
                else
                    low=k+1;
            }
        }
        cout<<a<<" "<<b<<endl;
    }
    return 0;
}