ZOJ 3665 Yukari's Birthday 第37届ACM/ICPC长春赛区现场赛K题 （水题，枚举，二分）

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

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Author: WU, Zejun
Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest

 

 

长春现场赛时，第一道题就是做的这题。

就是枚举r,然后用二分来求对应的k.

由于题目上k>=2，所以r肯定不会很大，二分就可以了。

#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<iostream>

#include<math.h>

using namespace std;

long long solve(int r,long long sum)

{

    long long left=2,right=(long long)sqrt((double)sum);

    while(left<=right)

    {

        long long mid=(left+right)/2;

        double mm=(pow(1.0*mid,r+1)-mid)/(mid-1);

        if(mm>((double)sum+1e-2))

        {

            right=mid-1;

            continue;

        }

        long long temp=0;

        long long tt=1;

        for(int i=0;i<r;i++)

        {

            tt*=mid;

            temp+=tt;

        }

        if(temp==sum)return mid;

        else if(temp<sum)left=mid+1;

        else right=mid-1;

    }

    return 0;

}



int main()

{

    long long n;

    long long ans,ansk;

    int ansr;

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

    {

        ans=n-1;

        ansr=1;

        ansk=n-1;

        for(int r=2;r<=40;r++)

        {

            long long temp=solve(r,n);

            //printf("%d\n",temp);

            if(temp==0)continue;

            if(r*temp<ans)

            {

                ans=r*temp;

                ansr=r;

                ansk=temp;

            }

            else if(r*temp==ans&&r<ansr)

            {

                ansr=r;

                ansk=temp;

            }

        }

        for(int r=2;r<=40;r++)

        {

            long long temp=solve(r,n-1);

            if(temp==0)continue;

            if(r*temp<ans)

            {

                ans=r*temp;

                ansr=r;

                ansk=temp;

            }

            else if(r*temp==ans&&r<ansr)

            {

                ansr=r;

                ansk=temp;

            }

        }

        printf("%d %lld\n",ansr,ansk);

    }

    return 0;

}