HDU 4430 Yukari's Birthday (二分+枚举)

题意：给定一个n（18 ≤ n ≤ 10^12），一个等比数列k + k^2 + .......+ k^r = n 或者 = n-1，求出最小的k*r，如果最小的不唯一，则取r更小的

分析：两个未知数，r,k，很明显，r的范围只有几十而已，所以枚举r；k的范围很大，需要二分...................

二分k的上界依情况而定 ： pow(n,1.0/i)；

 

#include <iostream>

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <string>

#include <vector>

#include <set>

#include <queue>

#include <stack>

#include <climits>//形如INT_MAX一类的

#define MAX 100005

#define INF 0x7FFFFFFFFFFFFFFF

#define REP(i,s,t) for(int i=(s);i<=(t);++i)

#define ll __int64

#define mem(a,b) memset(a,b,sizeof(a))

#define mp(a,b) make_pair(a,b)

#define L(x) x<<1

#define R(x) x<<1|1

# define eps 1e-5

//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂

using namespace std;

__int64 n;



struct node {

    __int64 k,ans;

    int r;

}minn;



__int64 pow2(__int64 mid,int r) {

    __int64 t = mid;

    for(int i=0; i<r-1; i++) {

        mid *= t;

    }

    return mid;

}



__int64 search(int r,__int64 tmp,__int64 high) {

    __int64 low = 2,mid;

    while(low <= high ) {

        mid = (low + high) >> 1;

        __int64 cal = (pow2(mid,(r+1)) - mid) / (mid-1);

        //cout << "k: " << mid << " r: " << r << " cal: " << cal << endl;

        if(cal > tmp) {

            high = mid - 1;

        } else if(cal < tmp) {

            low = mid + 1;

        } else {

            return mid;

        }

    }

    return -1;

}



int main(){

    while(cin >> n) {

        int high = log(n) / log(2);

        minn.ans = n-1;

        minn.r = 1;

        minn.k = n-1;

        for(int i=2; i<=high; i++) {

            int tmp = pow(n,1.0/i);

            __int64 tmp1 = search(i,n,tmp);

            __int64 tmp2 = search(i,n-1,tmp);

           // cout << "i: " << i << " tmp: " << tmp << " tmp1: " << tmp1 << " tmp2: " << tmp2 << endl;

            if(tmp1 != -1 && i * tmp1 <= minn.ans) {

                if(tmp1 * i == minn.ans && i < minn.r) {

                    minn.r = i; minn.k = tmp1;

                } else if(i * tmp1 < minn.ans){

                    minn.ans = i * tmp1; minn.r = i; minn.k = tmp1;

                }

            }

            if(tmp2 != -1 && i * tmp2 <= minn.ans) {

                if(tmp2 * i == minn.ans && i < minn.r) {

                    minn.r = i; minn.k = tmp2;

                } else if(i * tmp2 < minn.ans){

                    minn.ans = i * tmp2; minn.r = i; minn.k = tmp2;

                }

            }

        }

        printf("%d %I64d\n",minn.r,minn.k);

    }

    return 0;

}