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题目大意：给你你个n， 圆心放一个蜡烛，之后每一圈放k^i个，i = 1，2，3，4，5…….。求出来圈数r，第一圈的k的乘积最小，如果相同的时候求r最小。

10000组数据暴力果断TLE，后来发现r很小，可以暴力枚举r然后二分k。

Yukari‘s Birthday

18
111
1111

1 17
2 10
3 10

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <time.h>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
///#define LL long long
#define LL __int64
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007


using namespace std;

const int maxn = 1010;

LL Get(int x, LL n, LL ns)
{
    LL s = 1LL;
    for(int i = 1; i <= x; i++)
    {
        s *= n;
        if(n*(1LL-s)/(1LL-n) > ns) return -1;
    }
    return n*(1LL-s)/(1LL-n);
}


LL Find(int x, LL n)
{
    LL l = 2LL;
    LL r = (LL)pow(n, 1.0/x);
    while(l <= r)
    {
        LL mid = (l+r)>>1;
        LL x1 = Get(x, mid, n);
        if(x1 == -1LL)
        {
            r = mid-1LL;
            continue;
        }
        if(x1 == n || x1 == n-1) return mid;
        else if(x1 > n-1LL) r = mid-1LL;
        else if(x1 < n-1LL) l = mid+1LL;
    }
    return -1;
}

int main()
{
    LL n;
    while(~scanf("%I64d",&n))
    {
        LL Min = n-1;
        LL xi = 1LL;
        LL yi = n-1LL;
        LL ans = 0LL;
        while((1LL<<(ans)) < n) ans++;
        for(LL i = 1LL; i <= ans; i++)
        {
            LL x = Find(i, n);
            if(x == -1) continue;
            if(Min > x*i)
            {
                Min = x*i;
                xi = i;
                yi = x;
            }
            else if(Min == x*i)
            {
                if(xi > i)
                {
                    xi = i;
                    yi = x;
                }
            }
        }
        cout<<xi<<" "<<yi<<endl;
    }
    return 0;
}

HDU 4430 Yukari's Birthday(枚举+二分)

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原文地址：http://blog.csdn.net/xu12110501127/article/details/41516539