## ZOJ 3665 Yukari's Birthday 第37届ACM/ICPC长春赛区现场赛K题 （水题，枚举，二分）

Yukari's Birthday

Time Limit: 2 Seconds      Memory Limit: 32768 KB

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 ≤ ir. 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


Author: WU, Zejun
Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
long long solve(int r,long long sum)
{
long long left=2,right=(long long)sqrt((double)sum);
while(left<=right)
{
long long mid=(left+right)/2;
double mm=(pow(1.0*mid,r+1)-mid)/(mid-1);
if(mm>((double)sum+1e-2))
{
right=mid-1;
continue;
}
long long temp=0;
long long tt=1;
for(int i=0;i<r;i++)
{
tt*=mid;
temp+=tt;
}
if(temp==sum)return mid;
else if(temp<sum)left=mid+1;
else right=mid-1;
}
return 0;
}

int main()
{
long long n;
long long ans,ansk;
int ansr;
while(scanf("%lld",&n)!=EOF)
{
ans=n-1;
ansr=1;
ansk=n-1;
for(int r=2;r<=40;r++)
{
long long temp=solve(r,n);
//printf("%d\n",temp);
if(temp==0)continue;
if(r*temp<ans)
{
ans=r*temp;
ansr=r;
ansk=temp;
}
else if(r*temp==ans&&r<ansr)
{
ansr=r;
ansk=temp;
}
}
for(int r=2;r<=40;r++)
{
long long temp=solve(r,n-1);
if(temp==0)continue;
if(r*temp<ans)
{
ans=r*temp;
ansr=r;
ansk=temp;
}
else if(r*temp==ans&&r<ansr)
{
ansr=r;
ansk=temp;
}
}
printf("%d %lld\n",ansr,ansk);
}
return 0;
}

posted on 2012-10-16 12:29  kuangbin  阅读(1573)  评论(7编辑  收藏  举报

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