HDU 4430 & ZOJ 3665 Yukari's Birthday(二分法+枚举)

主题链接：

HDU:

pid=4430" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=4430

ZJU:

problemId=4888" target="_blank">http://acm.zju.edu.cn/onlinejudge/showProblem.do?

problemId=4888

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

 

Source

2012 Asia ChangChun Regional Contest

题意：

要在一个蛋糕上放置 n 根蜡烛，摆成 r 个同心圆，每一个同心圆的蜡烛数为 k ^ i ，中间的圆心能够放一根或者不放，使得 r * k 最小，若有多个答案输出 r 最小的那个。

PS:

由于r是非常小的 ！

枚举r查找k。

代码例如以下：（HDU，ZOJ上把64位换为long long就OK啦……）

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
//typedef long long LL;
typedef __int64 LL;
#define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
LL n;
LL findd(LL m)
{
    LL l, r, mid;
    l = 2;
    r = n;
    while(l <= r)
    {
        mid = (l+r)/2;
        LL sum = 0, tt = 1;
        for(LL i = 1; i <= m; i++)
        {
            if(n/tt < mid)//注意可能溢出用除法推断一下
            {
                //tt*mid > n
                sum = n+1;
                break;
            }
            tt*=mid;
            sum += tt;
//            if(sum > n)//防止溢出
//                break;
        }
        if(sum == n-1 || sum == n)
        {
            return mid;
        }
        if(sum < n-1)
        {
            l = mid+1;
        }
        else if(sum > n)
        {
            r = mid-1;
        }
    }
    return -1;//没有符合的
}

int main()
{
    LL r, k, rr, kk;
    while(~scanf("%I64d",&n))
    {
        rr = r = 1;
        kk = k = n-1;
        for(LL i = 2; i <= 64; i++)
        {
            LL tt = findd(i);
//            if(i >= n)
//                break;
//            printf("tt:%I64d>>>%I64d\n",i,tt);
            if(i*tt < rr*kk && tt != -1)
            {
                r = i;
                k = tt;
                rr = i;
                kk = tt;
            }
        }
        printf("%I64d %I64d\n",r,k);
    }
    return 0;
}

/*
18
111
1111
1022
8190
134217726
34359738366
68719476734
*/

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