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

枚举r,然后用二分来求对应的k.
因为题目上k>=2，所以r肯定不会很大，二分就可以了。
需要注意的是二分时初始右边界不能太大，不然会导致计算过程超出long long
还有就是提交时输入输出在zoj上要要用%lld，而hdu要用%I64d，不然会Wrong Answer

代码：
[cpp]?
#include?
#include?
#include?
#include?
long long powLL(long long a,int b){?
??? long long res=1;?
??? for(int i=0;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=i;?
??????????????????????? k=mm;?
??????????????????? }?
??????????????????? break;?
??????????????? }?
??????????????? else if(ans>n){?
??????????????????? rr=mm-1;?
??????????????? }?
??????????????? else {?
??????????????????? ll=mm+1;?
??????????????? }?
??????????? }?
??????? }?
??????? printf("%lld %lld\n",r,k);?
??? }?
??? return 0;?
}?