hdu 4430 Yukari's Birthday

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6314    Accepted Submission(s): 1520

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 <cmath>#include <algorithm>#define LL long longusing namespace std;LL mypow(LL x,LL y){    LL ans = 1;    while(y){        if(y&1) ans *= x;        y >>= 1;        x *= x;    }    return ans;}int main() {    LL n,r,k,lt,rt,mid;    while(~scanf("%I64d",&n)){        r = 1;        k = n-1;        for(int i = 2; i <= 41; i++){            lt = 2;            rt = (LL)pow(n,1.0/i);            while(lt <= rt){                mid = (lt+rt)>>1;                LL sum = (mid-mypow(mid,i+1))/(1-mid); //二分。根据等比数列，可知x(1-xn) = k*(1-x)                if(sum == n||sum == n-1){                    if(i*mid < r*k){                        r = i;                        k = mid;                    }                    break;                }else if(sum < n) lt = mid+1;                else rt = mid-1;            }        }        printf("%I64d %I64d\n",r,k);    }    return 0;}