首页 > ACM题库 > HDU-杭电 > hdu 4430-yukari’s birthday-分治-[解题报告]hoj
2015
07-16

hdu 4430-yukari’s birthday-分治-[解题报告]hoj

Yukari’s Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2549    Accepted Submission(s): 522



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
 
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef __int64 LL;
using namespace std;

const int MAX=1000+10;
LL n;

LL cal(LL k,LL r){
    LL sum=1,ans=0;
    for(int i=1;i<=r;++i){
    	if(n/sum<k)return n+1; 
        sum=sum*k;//sum*k可能会溢出 
        ans+=sum;
        if(ans>n)return ans;//ans大于n直接返回 
    } 
    return ans;
}

LL search(LL i){
    LL l=2,r=n;
    while(l<=r){
        LL mid=(l+r)>>1;
        LL sum=cal(mid,i);
        if(sum == n-1 || sum == n)return mid;
        if(sum<n-1)l=mid+1;
        else r=mid-1;
    }
    return -1;
}

int main(){
    while(~scanf("%I64d",&n)){
        LL a=1,b=n-1;
        for(LL i=2;i<=60;++i){//枚举r然后对k进行二分查找,因为k>?=2,2^60>10^12,所以i只要枚举到60即可 
            LL k=search(i);
            if(k != -1 && i*k<a*b){a=i,b=k;}
        }
        printf("%I64d %I64d\n",a,b);
    }
    return 0;
}

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