主题链接：
HDU: pid=4430" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=4430
ZJU: problemId=4888" target="_blank">http://acm.zju.edu.cn/onlinejudge/showProblem.do? problemId=4888
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.
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.
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
题意：
要在一个蛋糕上放置 n 根蜡烛，摆成 r 个同心圆，每一个同心圆的蜡烛数为 k ^ i ，中间的圆心能够放一根或者不放，使得 r * k 最小，若有多个答案输出 r 最小的那个。
PS:
由于r是非常小的 ！
枚举r查找k。
代码例如以下：（HDU，ZOJ上把64位换为long long就OK啦……）
#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
//typedef long long LL;
typedef __int64 LL;
#define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
LL n;
LL findd(LL m)
{
LL l, r, mid;
l = 2;
r = n;
while(l <= r)
{
mid = (l+r)/2;
LL sum = 0, tt = 1;
for(LL i = 1; i <= m; i++)
{
if(n/tt < mid)//注意可能溢出用除法推断一下
{
//tt*mid > n
sum = n+1;
break;
}
tt*=mid;
sum += tt;
// if(sum > n)//防止溢出
// break;
}
if(sum == n-1 || sum == n)
{
return mid;
}
if(sum < n-1)
{
l = mid+1;
}
else if(sum > n)
{
r = mid-1;
}
}
return -1;//没有符合的
} int main()
{
LL r, k, rr, kk;
while(~scanf("%I64d",&n))
{
rr = r = 1;
kk = k = n-1;
for(LL i = 2; i <= 64; i++)
{
LL tt = findd(i);
// if(i >= n)
// break;
// printf("tt:%I64d>>>%I64d\n",i,tt);
if(i*tt < rr*kk && tt != -1)
{
r = i;
k = tt;
rr = i;
kk = tt;
}
}
printf("%I64d %I64d\n",r,k);
}
return 0;
} /*
18
111
1111
1022
8190
134217726
34359738366
68719476734
*/
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