ZOJ 3665 Yukari's Birthday

ZOJ Problem Set - 3665

Yukari's Birthday

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Time Limit: 2 Seconds      Memory Limit: 32768 KB

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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

//枚举 r ，二分 k

#include <iostream>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

using namespace std;

#define __int64 long long  //zoj上的特殊需要哈

__int64 rc[2];

__int64 n;

int ok(int r,__int64 k)

{

    __int64 sum=0;

    for(int i=1;i<=r;i++)

    {

        sum+=(__int64)pow(1.0*k,1.0*i);

        if(sum>n) return 2;

    }

    if(sum==n||sum+1==n) return 3;

     return 1;

}

__int64 bf(int id,__int64 r) 

{

    __int64 l=2;

    __int64 m;

    int flag;

    while(l<=r)

    {

        m=(l+r)>>1;

        flag=ok(id,m);

        if(flag==1) l=m+1;

        else if(flag==2) r=m-1;

             else return m;

    }

    return 0;

}

void solve()

{

    __int64 id;

    int i;

    __int64 r=n;



    for(i=2;;i++) //枚举 r

    {

        r=(__int64)(pow(n*1.0,1.0/i)); 计算 k 的二分范围

        if(r==1) break;

        id=bf(i,r);,

        if(id)

        {

            if(rc[0]*rc[1]>i*id)

            {

                rc[0]=i;

                rc[1]=id;

            }

        }

    }

}

int main()

{

    while(scanf("%lld",&n)!=EOF)

    {

        rc[0]=1;rc[1]=n-1;

        solve();

        printf("%lld %lld\n",rc[0],rc[1]);

    }



    return 0;

}