hdu 4430 Yukari's Birthday 枚举+二分

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<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;
}

补充：软件开发 , C++ ,