HDU 4430 Yukari's Birthday

Yukari's Birthday

Time Limit: 6000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4430
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

2012 Asia ChangChun Regional Contest

 

解题：二分。根据等比数列，可知x(1-xⁿ) = k*(1-x)

 

n的范围比较小，枚举n再二分x即可。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using 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);
                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;
}