HDOJ 4430 Yukari's Birthday

C++高精度问题太蛋疼了....

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2891    Accepted Submission(s): 604

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

181111111

 

Sample Output

1 172 103 10

 

Source

2012 Asia ChangChun Regional Contest

 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long int LL;LL n;LL check(LL r){    LL low=2,high=n,mid,ret=-1;    while(low<=high)    {        mid=(low+high)/2;        if(((LL)(pow(mid*1.,r-1.))>n||(LL)(pow(mid*1.,r-1.))<0)           ||(LL)(pow(mid*1.,r*1.))>n||(LL)(pow(mid*1.,r*1.))<0)        {            high=mid-1;            continue;        }        LL temp=mid*(1-(LL)(pow(mid*1.,r*1.)))/(1-mid);        if(temp<=n-1||temp<=n)        {            if(temp==n||temp==n-1) ret=mid;            low=mid+1;        }        else        {            high=mid-1;        }    }    return ret;}int main(){while(scanf("%I64d",&n)!=EOF){    LL R=1,K=n-1,ans=(1LL<<60);    for(LL r=2;r<40;r++)    {        LL k=check(r);        if(k==-1) continue;        if(ans>r*k)        {            R=r; K=k;        }    }    printf("%I64d %I64d\n",R,K);}    return 0;}