HDU 4430 Yukari's Birthday（枚举+二分）

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6103    Accepted Submission(s): 1438


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

题意：有n个蜡烛，插成同心圆，中心可插可不插，一共r圈，第i圈插k^i个。求r*k最小的一组

解决：先枚举r，再二分k

细节：数据大小，二分注意判断更改左端右端。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define ll long long

ll n;

ll bin(ll s)
{
ll l = 2,r = n,mid,i;
while(l<=r)
{
mid = (l+r)/2;
ll sum = 1,ans = 0;;
for(i = 1; i<=s; i++)
{
if(n/sum<mid)
{
ans = n+1;
break;
}
sum*=mid;
ans+=sum;
if(ans>n)//放置溢出
break;
}
if(ans == n || ans == n-1) return mid;
else if(ans<n-1) l = mid+1;
else r = mid-1;
}
return -1;
}

int main()
{
ll s,k,l,r,mid,i;
while(~scanf("%I64d",&n))
{
l = 1,r = n-1;
for(i = 2; i<=45; i++)
{
k = bin(i);
if(k!=-1 && i*k<l*r)
{
l = i,r = k;
}
}
printf("%I64d %I64d\n",l,r);
}

return 0;
}