HDU 4430

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6849    Accepted Submission(s): 1662


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
 

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题意:给n跟蜡烛,第0层最多放一根蜡烛,第i层要放k^i个蜡烛(共r层,k>=2),要把所有蜡烛放完,求k,r,使k*r最小

2^40>1e12,所以r最大不超过40,最后一层的蜡烛数为k^r<=n,k


#pragma comment(linker,"/STACK:1024000000,1024000000")
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair PII;

#define pi acos(-1.0)
#define eps 1e-10
#define pf printf
#define sf scanf
#define lson rt<<1,l,mi
#define rson rt<<1|1,mi+1,r
#define root 1,1,n
#define e tree[rt]
#define _s second
#define _f first
#define all(x) (x).begin,(x).end
#define mem(i,a) memset(i,a,sizeof i)
#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
#define mi ((l+r)>>1)
#define sqr(x) ((x)*(x))

const int inf=0x3f3f3f3f;
const int mod=1e9+7;
ll m,k,R;

ll quick(ll x,ll y)
{
    ll ans=1;
    while(y)
    {
        if(y&1)ans=ans*x;
        y>>=1;
        x=x*x;
    }
    return ans;
}

void solve(ll x,int y)
{
    int l=2,r=pow(x,1.0/y)+1;//L要从2开始,除数可能为0
    while(l>1;
        ll t=(quick(mid,y)-1)/(mid-1)*mid;
        if(t==m||t==m-1)
        {
            if(k*R>mid*y)
                k=mid,R=y;
            return ;
        }
        else if(t>m)
            r=mid-1;
        else
            l=mid+1;
    }
}

int main()
{
    while(~sf("%I64d",&m))
    {
        k=m-1,R=1;
        for(int i=2;i<41;i++)
            solve(m,i);
        pf("%I64d %I64d\n",R,k);
    }
    return 0;
}





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