HDOJ 4937 Lucky Number

当进制转换后所剩下的为数较少时（2位，3位），对应的base都比较大，可以用数学的方法计算出来。

预处理掉转换后位数为3位后，base就小于n的3次方了，可以暴力计算。。。。

Lucky Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 521    Accepted Submission(s): 150

Problem Description

“Ladies and Gentlemen, It’s show time! ”

“A thief is a creative artist who takes his prey in style... But a detective is nothing more than a critic, who follows our footsteps...”

Love_Kid is crazy about Kaito Kid , he think 3(because 3 is the sum of 1 and 2), 4, 5, 6 are his lucky numbers and all others are not. 

Now he finds out a way that he can represent a number through decimal representation in another numeral system to get a number only contain 3, 4, 5, 6. 

For example, given a number 19, you can represent it as 34 with base 5, so we can call 5 is a lucky base for number 19. 

Now he will give you a long number n(1<=n<=1e12), please help him to find out how many lucky bases for that number. 

If there are infinite such base, just print out -1.

 

Input

There are multiply test cases.

The first line contains an integer T(T<=200), indicates the number of cases. 

For every test case, there is a number n indicates the number.

 

Output

For each test case, output “Case #k: ”first, k is the case number, from 1 to T , then, output a line with one integer, the answer to the query.

 

Sample Input

   
   
   
   

    
    
    
    2
10
19
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 0
Case #2: 1


    
    
    
    
 
     
 
      Hint 
     
10 shown in hexadecimal number system is another letter different from ‘0’-‘9’, we can represent it as ‘A’, and you can extend to other cases. 
    
    
    
    
     
   
   
   
   

 

Author

UESTC

 

Source

2014 Multi-University Training Contest 7

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <cmath>

using namespace std;

typedef long long int LL;

LL n;
set<LL> ans;
bool change(LL x,LL base)
{
    bool flag=true;
    while(x)
    {
        LL temp=x%base;
        x/=base;
        if(temp>=3LL&&temp<=6LL) continue;
        else
        {
            flag=false;
            break;
        }
    }
    return flag;
}

int main()
{
    int T_T,cas=1;
    scanf("%d",&T_T);
    while(T_T--)
    {
        ans.clear();
        cin>>n;
        if(n>=3LL&&n<=6LL)
        {
           cout<<"Case #"<<cas++<<": -1"<<endl; continue;
        }

        for(LL a=3;a<=6;a++)
        {
            for(LL b=3;b<=6;b++)
            {
                LL limit=max(a,b);
                if( (n-b)%a == 0 )
                {
                    if( (n-b)/a > limit)
                    {
                        ans.insert((n-b)/a);
                    }
                }
            }
        }

        for(LL a=3;a<=6;a++)
        {
            for(LL b=3;b<=6;b++)
            {
                for(LL c=3;c<=6;c++)
                {
                    LL C=c-n;
                    if(b*b >= 4LL*a*C )
                    {
                        LL deta=sqrt(b*b-4LL*a*C);
                        LL base1=(-b+deta)/(2*a);
                        LL base2=(-b-deta)/(2*a);
                        LL limit=max(a,max(b,c));
                        if(a*base1*base1+b*base1+c==n && base1>limit)
                        {
                            ans.insert(base1);
                        }
                        if(a*base2*base2+b*base2+c==n && base2>limit)
                        {
                            ans.insert(base2);
                        }
                    }
                }
            }
        }

        for(LL i=4;i*i*i<=n;i++)
        {
            if(change(n,i))
            {
                ans.insert(i);
            }
        }
        cout<<"Case #"<<cas++<<": "<<ans.size()<<endl;
    }
    return 0;
}