hdu 4937 2014 Multi-University Training Contest 7 1003

Lucky Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 664    Accepted Submission(s): 194

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

2 10 19

 

Sample Output

Case #1: 0 Case #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

 

Recommend

We have carefully selected several similar problems for you:  4944 4943 4942 4941 4940 

 

题意、题解,转自：

http://blog.csdn.net/a601025382s/article/details/38517783

 

题意：

我们将3,4,5,6认为是幸运数字。给定一个十进制数n。现在可以讲起任意转换成其他进制，但转换后的数必须是由3,4,5,6构成的，而这个进制称为幸运进制。问有多少个幸运进制。若有无数个，则输出-1。例如19在5进制下是34，所以5是幸运进制。

题解：

先考虑特殊情况，所情况下会有无穷个？只有n=3,4,5,6的时候，因为这几个数在大于n的进制下都是他本身。。注意特殊情况不包括33,343这些（我一开始就死在这里了，wa了三次）。因为33在34进制下就不是33了（类似于10在16进制下就是A了）。

我们知道n=a0+a1*x+a2*x^2+...，其中x为进制。由于n达到1e12，所以我们分情况讨论。

1）a0形式，我们已经在特殊情况中指出，只有无穷个的时候才会符合条件

2）a0+a1*x形式，枚举a0,a1，我们判断(n-a0)是否能被a1整除，以及x是否大于max(a0,a1)即可。

3）a0+a1*x+a2*x^2，我们枚举a0,a1,a2，那么就相当于解一元二次方程。判断是否有整数解，是否整数解x>max(a0,a1,a2)即可。

4）不在上述三种形式内的，那么进制x最大也不会x^3>n，不然就会变成上述三种的形式。我们就可以枚举进制然后判断是否为幸运进制了。由于x^3<=n，所以复杂度只有1e4。

注意：就是上述的特殊情况，死的惨惨的。。

代码：

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>

#define N 100005
#define M 10005
#define mod 1000000007
#define mod2 100000000
#define ll long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

int T;
int f[N];
ll n;

void ini()
{
    memset(f,0,sizeof(f));
    int i,j;
    int te,yu;
    for(i=1;i<=M;i++){
        for(j=4;j<=10000;j++){
            te=i;
            int flag=1;
            while(te){
                yu=te%j;
                if(yu!=3 && yu!=4 && yu!=5 && yu!=6){
                    flag=0;break;
                }
                te/=j;
            }
            if(flag==1) f[i]++;
        }
    }
    f[3]=f[4]=f[5]=f[6]=-1;

   // for(i=1;i<=M;i++){
    //    printf(" i=%d f=%d
",i,f[i]);
    //}
}

int main()
{
    ll ans;
    ll j,i,k;
    ll a,b,c,d;
    ll base;
    //freopen("data.in","r",stdin);
   // ini();
    scanf("%d",&T);
    for(int cnt=1;cnt<=T;cnt++)
    {
        ans=0;
        printf("Case #%d: ",cnt);
        scanf("%I64d",&n);
        if(n>=3 && n<=6){
            printf("-1
");continue;
        }

        for(i=3;i<=6;i++){
            for(j=3;j<=6;j++){
                if( (n-i)%j==0 && (n-i)/j >max(i,j) ) ans++;
            }
        }
      //  printf(" %I64d
",ans);

        for(i=3;i<=6;i++){
            for(j=3;j<=6;j++){
                for(k=3;k<=6;k++){
                    a=i;b=j;c=k-n;
                    ll te=sqrt(b*b-4*a*c+0.5);
                    if(te*te!=b*b-4*a*c) continue;
                    if(te<b) continue;
                    d=(te-b);
                    if(d%(2*a)==0){
                        base=d/2/a;
                        if(base>max(max(i,j),k))ans++;
                    }
                }
            }
        }

      //  printf("  %I64d
",ans);

        //printf("%I64d
",ans);

        for(j=4;j*j*j<=n;j++){
            ll te=n;
            int flag=1;
                while(te){
                    ll yu=te%j;
                    if(yu!=3 && yu!=4 && yu!=5 && yu!=6){
                        flag=0;break;
                    }
                    te/=j;
                }
            if(flag==1) ans++;
        }
        printf("%I64d
",ans);
       // }


    }
    return 0;
}