HDU4430 Yukari's Birthday 二分

柔光的暖阳◎ 2022-05-27 06:24 140阅读 0赞

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

k>=2，所以r毛估估，应该是不会超过45的，枚举一下r，然后二分找k就可以了。

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<algorithm>
    #include<map>
    #include<queue>
    #include<vector>
    using namespace std;
    long long n;
    long long findk(long long x)
    {//二分 
    	long long l=2,r=n,mid,i,s,sum;
    	while(l<=r)
    	{
    		mid=(l+r)/2;
    		s=1;sum=0;
    		for(i=1;i<=x;i++)
    		{
    			if(n/s<mid)
    			{
    				sum=n+1;
    				break;
    			}
    			s*=mid;//k^i 
    			sum+=s;//和 
    			if(sum>n)break;//太大了 
    		}
    		if(sum==n||sum+1==n)//圆心可以放or不放 
    		return mid;
    		else if(sum<n-1)
    		l=mid+1;
    		else
    		r=mid-1;
    	}
    	return -1; 
    }
    int main()
    {
    	long long i,j,r,k,s,o,p;
    	while(scanf("%lld",&n)!=EOF)
    	{
    		r=1;k=n-1;
    		s=n-1;
    		for(i=2;i<=45;i++)//枚举r
    		{
    			o=findk(i);//new k
    			p=o*i;//k*r
    			if(o!=-1&&p<s)
    			{
    				s=p;
    				r=i;k=o;//更新 
    			}
    		} 
    		printf("%lld %lld\n",r,k);
    	} 
    }