Hdu4937: lucky number

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: 0 case #2: 1 for a single digit, 3, 4, 5, 6 is obviously-1 for two places, solving the equation A * x + B = N for three places, connect to Great Wall A * x ^ 2 + B * x + c = n. Other values start from 4 hexadecimal enumeration.
#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 __int64int 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;}