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-x ⁿ) = k*(1-x)

 

n的范围比较小，枚举n再二分x即可。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 LL mypow(LL x,LL y){

18     LL ans = 1;

19     while(y){

20         if(y&1) ans *= x;

21         y >>= 1;

22         x *= x;

23     }

24     return ans;

25 }

26 int main() {

27     LL n,r,k,lt,rt,mid;

28     while(~scanf("%I64d",&n)){

29         r = 1;

30         k = n-1;

31         for(int i = 2; i <= 41; i++){

32             lt = 2;

33             rt = (LL)pow(n,1.0/i);

34             while(lt <= rt){

35                 mid = (lt+rt)>>1;

36                 LL sum = (mid-mypow(mid,i+1))/(1-mid);

37                 if(sum == n||sum == n-1){

38                     if(i*mid < r*k){

39                         r = i;

40                         k = mid;

41                     }

42                     break;

43                 }else if(sum < n) lt = mid+1;

44                 else rt = mid-1;

45             }

46         }

47         printf("%I64d %I64d\n",r,k);

48     }

49     return 0;

50 }

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