HDOJ 4937 Lucky Number

HDOJ 4937 Lucky Number

When there are only a few (two or three digits) after the hexadecimal conversion, the corresponding base is relatively large and can be calculated using a mathematical method.

After the number of digits after the pre-processing and conversion is 3, the base is less than the power of n, which can be brute force calculated ....

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.
InputThere 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.
OutputFor 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

21019

Sample Output

Case #1: 0Case #2: 1Hint10 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. 

AuthorUESTC
Source2014 Multi-University Training Contest 7

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     #include 
     
      using namespace std;typedef long long int LL;LL n;set
      
        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 #"<
       
         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 #"<