HDU - 4430 Yukari's Birthday(二分)

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Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6368    Accepted Submission(s): 1539

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

 

Source

2012 Asia ChangChun Regional Contest

 

Recommend

zhuyuanchen520

 

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题目大意：

给出蜡烛数n,打算摆成同心圆，要求每一层i的蜡烛数为k^i.要求给出使得r*k最小的r和k。其中r为同心圆的个数，k为第一层的蜡烛数。最中心选择可以插一个蜡烛或者不插。

思路：

r最多40，而k最多1000000。所以可以枚举r,再对k二分。注意讨论中心有无蜡烛两种情况。

附上AC代码：

#include

using namespace std;
typedef long long ll;
const int INF=0x7f7f7f;
ll ans,n,k;
int r;
//二分枚举k
int binary(int r,ll n){
    ll ln=2,rn=1000000,sum,tmp;
    while(ln<=rn){
        ll mid=(ln+rn)>>1;
        sum=1,tmp=1;
        for(int i=1;i<=r;i++){
            tmp*=mid;
            sum+=tmp;
            if(sum>n)break;
        }
        if(sum==n)return mid;
        else if(sum>n){
            rn=mid-1;
        }
        else ln=mid+1;
    }
    return -1;
}

int main(){
    ios::sync_with_stdio(false);
    while(cin>>n){
        ans=INF;
        r=1,k=n-1;
        //遍历r
        for(int i=1;i<=40;i++){
            ll tmp=binary(i,n);
            if(tmp!=-1){
                if(i*tmp