hdu 4937 Lucky Number 2014 Multi-University Training Contest 7

Lucky Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 211    Accepted Submission(s): 55

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.
 

 

Author

UESTC

 

Source

2014 Multi-University Training Contest 7

 

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比赛时怎么也没想到要特殊处理，就纠结于数据太大了

特殊处理只有1 2 3 位的情况 然后暴力枚举，数据量最大不会超过7000。

#include <cstdlib>#include <cstring>#include <cstdio>#include <iostream>#include <cmath>#include <algorithm>using namespace std;int ans;int main(){    int T;    long long n;    scanf("%d",&T);    int cas=1;    while(T--){        ans=0;        scanf("%I64d",&n);        if(n>=3&&n<=6){            printf("Case #%d: -1\n",cas++);            continue;        }        for(int i=3;i<=6;i++)            for(int j=3;j<=6;j++)                if(n>i&&(n-i)%j==0&&(n-i)/j>i&&(n-i)/j>j)ans++;        for(int i=3;i<=6;i++){            for(int j=3;j<=6;j++){                for(int k=3;k<=6;k++){                    long long p=j*j-4*i*(k-n);                    if(p>0){                        long long q=sqrt(p);                        if(q*q==p){                            if((-j+q)%(2*i)==0)                                   if((-j+q)/(2*i)>max(i,max(j,k)))ans++;                        }                    }                }            }        }        long long t,i;        for(i=4;i*i*i<=n;i++){            t=n;            while(t){                if(t%i<3||t%i>6)break;                t/=i;            }            if(!t)ans++;        }        printf("Case #%d: %d\n",cas++,ans);    }    return 0;}