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;
}