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,,sizeof(f));
     int i,j;
     int te,yu;
     for(i=;i<=M;i++){
         for(j=;j<=;j++){
             te=i;
             int flag=;
             while(te){
                 yu=te%j;
                 if(yu!= && yu!= && yu!= && yu!=){
                     flag=;break;
                 }
                 te/=j;
             }
             if(flag==) f[i]++;
         }
     }
     f[]=f[]=f[]=f[]=-;

    // for(i=1;i<=M;i++){
     //    printf(" i=%d f=%d\n",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=;cnt<=T;cnt++)
     {
         ans=;
         printf("Case #%d: ",cnt);
         scanf("%I64d",&n);
         if(n>= && n<=){
             printf("-1\n");continue;
         }

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

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

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

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

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

     }
     return ;
 }