当进制转换后所剩下的为数较少时（2位。3位），相应的base都比較大。能够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，能够暴力计算。
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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.
“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.
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: 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.
Author
UESTC
Source
#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;
}