hdu 4430 Yukari's Birthday

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1665 Accepted Submission(s): 341


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 1012.

[align=left]Output[/align]
For each test case, output r and k.

[align=left]Sample Input[/align]

18 111 1111

[align=left]Sample Output[/align]

1 17 2 10 3 10

// 因为 2^64-1 差不多是 10^18 所以 1+k^1+k^2+k^3+..+k^r == n ，r 不会很大
//所以我们可以枚举 r 接下来就是 二分查找 k 是不是存在
// 我们在计算  1+k^1+k^2+k^3+..+k^r 时 不要使用 等比公式 ，因为会超 long long
// 我们可以一项项计算 1+k^1+k^2+k^3+..+k^r ； 在计算 k ^ i 时 要判断 是不是超n
// 判断 k^i = k^(i-1) * k  > n 所以 改为除法 k^(i-1) > n / k
#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cmath>
#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std ;
#define LL long long
LL n , kk , rr , m ;

LL pow1( LL k , int r )
{
LL ans = 1 , sum = 0 ;
for( int i = 1 ; i <= r ;i++ ){
if( ans > n / k ) return n+1 ;
ans *= k ;
if( sum > n-ans ) return n+1 ;
sum += ans ;
}
return sum ;
}
void dfs( int r )
{
LL L , R , mid , sum ;
L = 2 ;
R = n-1 ;
while( R >= L )
{
mid = (R+L)/2 ;
sum = pow1(mid,r) ;
if( sum == n || sum == n-1 ) break ;

if( sum > n ) R = mid-1 ;
else L = mid+1 ;
}
LL k = mid ;
if( L > R ) return ;
if( sum == n || sum == n-1 )
{
if(k*r < rr*kk)
{
rr = r ; kk = k ;
}
return ;
}
}
int main()
{
int i ;
while( scanf("%I64d" , &n ) != EOF )
{
m = (LL)sqrt(n*1.0) ;
kk = n-1 ; rr = 1 ;
for( int j = 2 ; j <= 47 ;j++ )
dfs(j) ;
// hdu 用 lld 会wa ....
printf("%I64d %I64d\n" ,rr , kk ) ;
}
}