HDU4937：Lucky Number

“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.

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.

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.

2
10
19

Case #1: 0
Case #2: 1

对于一位的状况，3,4,5,6明显是-1
对于二位，解方程a*x+b=n
对于三位，接长城a*x^2+b*x+c=n
其他的从4进制开始枚举即可

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define w(x) while(x)
#define ll __int64

int main()
{
    int t,cas=1;
    ll n,tem,r,i,j,k,a,b,c,ans;
    scanf("%d",&t);
    w(t--)
    {
        ans=0;
        scanf("%I64d",&n);
        printf("Case #%d: ",cas++);
        if(n>2&&n<7)
        {
            printf("-1\n");
            continue;
        }
        up(i,3,6)
        up(j,3,6)
        if((n-i)%j==0 && (n-i)/j>max(i,j))
            ans++;
        up(i,3,6)
        up(j,3,6)
        up(k,3,6)
        {
            a=i,b=j,c=k-n;
            tem=(ll)sqrt(b*b-4*a*c+0.5);
            if(tem*tem!=b*b-4*a*c)
                continue;
            if((tem-b)%(2*a))
                continue;
            tem=(tem-b)/(2*a);
            if(tem>max(i,max(j,k)))
                ans++;
        }
        for(i=4;i*i*i<=n;i++)
        {
            tem=n;
            w(tem)
            {
                r=tem%i;
                if(r<3||r>6)
                    break;
                tem/=i;
            }
            if(!tem)
                ans++;
        }
        printf("%I64d\n",ans);
    }

    return 0;
}