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
18
111
1111
Sample Output
1 17
2 10
3 10
Source
2012 Asia ChangChun Regional Contest
枚舉r,然後用二分來求對應的k.
因為題目上k>=2，所以r肯定不會很大，二分就可以了。
需要注意的是二分時初始右邊界不能太大，不然會導致計算過程超出long long
還有就是提交時輸入輸出在zoj上要要用%lld，而hdu要用%I64d，不然會Wrong Answer
代碼：
[cpp]
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
long long powLL(long long a,int b){
long long res=1;
for(int i=0;i<b;i++)
res*=a;
return res;
}
int main(void){
long long n,r,k;
while(scanf("%lld",&n)!=EOF){
r=1;
k=n-1;
for(int i=2;i<=45;i++){
long long ll,rr,mm;
ll=2;
rr=(long long)pow(n,1.0/i);
while(ll<=rr){
mm=(long long)(ll+rr)/2;
long long 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("%lld %lld\n",r,k);
}
return 0;
}
