HDU 4430 二分查找

Yukari's Birthday
Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5071    Accepted Submission(s): 1199

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 k ⁱ 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 ≤ 10 ¹².

 

Output

For each test case, output r and k.

 

Sample Input

  
  

   
   18
111
1111
  
  

 

Sample Output

  
  

   
   1 17
2 10
3 10
  
  

 

题意：摆一圈一圈的蜡烛，中间的一个蜡烛可放可不放，不算圈数，层数为r，每层的蜡烛数为k^(1~r),k ≥ 2, 1 ≤ i ≤ r.给一个数，输出r*k最小的k和r。

由于当时老胡负责这个题目，所以把直接把他代码搬过来了：老胡传送门 老胡:http://www.cnblogs.com/pach/p/5759809.html

给出n，让n满足下列表达式：k^1+k^2+...+k^r=n. 且r*k要最小。(ps: And it's optional to place at most one candle at the center of the cake. 所以k^0，即1可有可无。但是这并不算一个圆，所以当n=30和n=31时，它们的r相等)  例如：2^1+2^2+2^3+2^4=30. n=30,r=4,k=2. (n=31时，r也为4，k也为2)  所以就有通解r=1,k=n-1. 这对于每个n都成立。不过我们得让r*k最小，就得都遍历一遍了。
因为2^0+2^1+2^2+...+2^40=2^41-1>10^12. 所以1<=r<=40. r不是很大，直接暴力。k通过二分来找。

当时考虑到一种情况就是当r>=2的时候，k的最大值是n-1，n最大为10^12,所以k=10^12-1,k^0+k^1+k^2<=n,所以二分查找k的时候直接从high=100w开始。（100w的平方为2的12次方），但是提交的时间并没有缩短反而增加，为什么？

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const long long inf=1000000000001;
int main()
{
    //freopen("in.txt","r",stdin);
    long long n;
    while(scanf("%I64d",&n)!=EOF)
    {
        int min_r;
        long long min_k,min_v=inf;
        for(int r=1; r<=40; r++)
        {
            long long low=1,high=n,k; //每一次r的更新，都要进行一次完整的二分查找
            while(1)
            {
                k=(low+high)/2;
                long long sum=0;
                bool flag=true;
                //此处是关键，由于数值很大，直接判断值是否超出n，如果用sum的方法判断又多了内存，其实在int64位后面多一位的相加直接可以忽略少一位的数值了
                for(int i=1; i<=r; i++)
                    if(pow(k*1.0,i)>n) //此处是判读那看k^i是否大于n，如果大于，就不用再求和了
                    {
                        sum=n+1;
                        flag=false;
                        break;
                    }
                if(flag)
                    for(int i=1; i<=r && sum<=n; i++)
                        sum+=(long long)pow(k*1.0,i);
                if(sum==n || sum==n-1) //因为1可有可无，所以满足其中一个条件即可
                    if(r*k<min_v)
                    {
                        min_r=r;
                        min_k=k;
                        min_v=r*k; //保存最小乘积
                    }
                if(sum>n)
                    high=k;//sum大了则从左边找
                else
                    low=k; 
                if(high-low==1)
                    break;
            }
        }
        printf("%d %I64d\n",min_r,min_k);
    }
    return 0;
}

还有就是这题右边的查找范围可能溢出，需要考虑。