HDOJ 4430 Yukari's Birthday

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5303    Accepted Submission(s): 1258

Problem Description

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

 

给你n个蜡烛要插在一个蛋糕上面，把蛋糕分成r层，每层上插的蜡烛数量是k的i次方，i是第i层，也就是求n==k^1+k^2+......k^r，以为中心可以插可以不插所以还有种可能就是

n-1==k^1+.....+k^r，仔细判断一下r在50以内切r大于等于2时，k在10000以内，因为要维护r*k最小，那么来暴力试一试，枚举r找k然后维护r*k最小，在找k中是等比，这里可以进行二分，暴力搞一下就可以过，但是要注意当r只能等于1时，k=n-1；这里要注意判断。

#include<vector>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<math.h>
#include<algorithm>
using namespace std;
#define ll long long
ll find_k(ll r,ll n)
{
    ll low=2;
    ll high=1000000;
    ll mid;
    while(low<=high)
    {
        mid=(low+high)/2;
        ll temp=mid;
        ll sum=0;
        for(int i=1;i<=r;i++)
        {
            sum+=temp;
            temp*=mid;
            if(sum>n)
                break;
        }
        if(sum==n)
            return mid;
        if(sum>n)
            high=mid-1;
        else
            low=mid+1;
    }
    return -1;
}
int main()
{
    ll n;
    while(scanf("%lld",&n)!=EOF)
    {
        ll ansr=-1,ansk;
        for(ll r=1;r<=50;r++)
        {
            ll tmpk=find_k(r,n);
            if(tmpk!=-1)
            {
                if(ansr==-1)
                {
                    ansr=r;
                    ansk=tmpk;
                }
                else
                {
                    if(ansr*ansk>tmpk*r)
                    {
                        ansr=r;
                        ansk=tmpk;
                    }
                }
            }
            tmpk=find_k(r,n-1);
            if(tmpk!=-1)
            {
                if(ansr==-1)
                {
                    ansr=r;
                    ansk=tmpk;
                }
                else
                {
                    if(ansr*ansk>tmpk*r)
                    {
                        ansr=r;
                        ansk=tmpk;
                    }
                }
            }
        }
        if(ansr==-1)
            printf("1 %lld\n",n-1);
        else
            printf("%lld %lld\n",ansr,ansk);
    }

return 0;
}