Yukari's birthday (precision processing) (Binary)

Yukari's birthday Time Limit: 12000/6000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 2910 accepted submission (s): 610


Problem descriptiontoday 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 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.
 
Inputthere are about 10,000 test cases. process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 1012.
 
Outputfor each test case, output R and K.
Sample Input

181111111

 
Sample output

1 172 103 10
# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # define LH _ int64 using namespace STD; LH powll (lh x, int y) {LH res = 1; for (INT I = 0; I <Y; I ++) {res * = x ;} return res;} int main () {lh n; lh r, K; while (~ Scanf ("% i64d", & N) {r = 1; k = n-1; lh ll, RR, mm; for (INT I = 2; I <= 45; I ++) {ll = 2; RR = (LH) Pow (n, 1.0/I); // specifies the upper bound while (LL <= RR) {mm = (LH) (LL + RR)/2; LH ans = (Mm-powll (mm, I + 1)/(1-mm ); if (ANS = n | ans = N-1) {if (I * mm <r * k) {r = I; k = mm;} break ;} else if (ANS> N) RR = mm-1; else LL = mm + 1 ;}} printf ("% i64d % i64d \ n", R, k );} return 0 ;}

Yukari's birthday (precision processing) (Binary)