HDU 4937 Lucky Number

Lucky Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1542    Accepted Submission(s): 453

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

21019

 

Sample Output

Case #1: 0Case #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.
 

题意：输入一个十进制数字n,用一个k进制数表示，每个数只能包含3 4 5 6，问这样的k有多少个

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <map>using namespace std;typedef long long ll;int main(){    int T;    scanf("%d",&T);    for(int ca=1;ca<=T;ca++)    {        ll n;        scanf("%I64d",&n);        printf("Case #%d: ",ca);        if(n>=3&&n<=6)//只有3 4 5 6是无限多种进制可以表示，10以上数字会用'A'表示        {            puts("-1");            continue;        }        int num=0;        for(int i=3;i<=6;i++)//两位的时候            for(int j=3;j<=6;j++)            {                if((n-j)%i==0 && (n-j)/i>i && (n-j)/i>j)                {                    num++;                }            }        for(int i=3;i<=6;i++)//三位组成的时候            for(int j=3;j<=6;j++)                for(int k=3;k<=6;k++)                {                    ll m=j*j-4*(k-n)*i;                    ll f=sqrt(m);                    if(f*f==m)                    {                        ll tt=(f-j)%(2*i);                        ll ff=(f-j)/(2*i);                        if(tt==0 && ff>i && ff>j && ff>k)                        {                            num++;                        }                    }                }        for(int i=4;i<=10000;i++)//大于等于四位的时候，此时进制数最大不会超过10000        {            ll t=n;            int flag=0;            int cnt=0;            while(t)            {                ll f=t%i;                if(f<3 || f>6)                {                    flag=1;                    break;                }                t/=i;                cnt++;//判断有几位，小于4位前面已经出现过，会重复            }            if(flag==0 && cnt>3)            {                num++;            }        }        printf("%d\n",num);    }    return 0;}