hdu 4937 2014 Multi-University Training Contest 7 1003

幸运数字与进制转换

最新推荐文章于 2020-08-02 20:54:41 发布

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最新推荐文章于 2020-08-02 20:54:41 发布

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原文链接：http://www.cnblogs.com/njczy2010/p/3909755.html

本文探讨了在特定幸运数字（3, 4, 5, 6）组成的数系中，如何将一个十进制数转换为幸运进制，并求解这样的幸运进制数量。文章详细介绍了解题思路与步骤，包括特殊情形处理、不同进制转换情况的分析等，最终通过代码实现了解决方案。

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

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题意、题解,转自：

http://blog.youkuaiyun.com/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：

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 
  9 #define N 100005
 10 #define M 10005
 11 #define mod 1000000007
 12 #define mod2 100000000
 13 #define ll long long
 14 #define maxi(a,b) (a)>(b)? (a) : (b)
 15 #define mini(a,b) (a)<(b)? (a) : (b)
 16 
 17 using namespace std;
 18 
 19 int T;
 20 int f[N];
 21 ll n;
 22 
 23 void ini()
 24 {
 25     memset(f,0,sizeof(f));
 26     int i,j;
 27     int te,yu;
 28     for(i=1;i<=M;i++){
 29         for(j=4;j<=10000;j++){
 30             te=i;
 31             int flag=1;
 32             while(te){
 33                 yu=te%j;
 34                 if(yu!=3 && yu!=4 && yu!=5 && yu!=6){
 35                     flag=0;break;
 36                 }
 37                 te/=j;
 38             }
 39             if(flag==1) f[i]++;
 40         }
 41     }
 42     f[3]=f[4]=f[5]=f[6]=-1;
 43 
 44    // for(i=1;i<=M;i++){
 45     //    printf(" i=%d f=%d\n",i,f[i]);
 46     //}
 47 }
 48 
 49 int main()
 50 {
 51     ll ans;
 52     ll j,i,k;
 53     ll a,b,c,d;
 54     ll base;
 55     //freopen("data.in","r",stdin);
 56    // ini();
 57     scanf("%d",&T);
 58     for(int cnt=1;cnt<=T;cnt++)
 59     {
 60         ans=0;
 61         printf("Case #%d: ",cnt);
 62         scanf("%I64d",&n);
 63         if(n>=3 && n<=6){
 64             printf("-1\n");continue;
 65         }
 66 
 67         for(i=3;i<=6;i++){
 68             for(j=3;j<=6;j++){
 69                 if( (n-i)%j==0 && (n-i)/j >max(i,j) ) ans++;
 70             }
 71         }
 72       //  printf(" %I64d\n",ans);
 73 
 74         for(i=3;i<=6;i++){
 75             for(j=3;j<=6;j++){
 76                 for(k=3;k<=6;k++){
 77                     a=i;b=j;c=k-n;
 78                     ll te=sqrt(b*b-4*a*c+0.5);
 79                     if(te*te!=b*b-4*a*c) continue;
 80                     if(te<b) continue;
 81                     d=(te-b);
 82                     if(d%(2*a)==0){
 83                         base=d/2/a;
 84                         if(base>max(max(i,j),k))ans++;
 85                     }
 86                 }
 87             }
 88         }
 89 
 90       //  printf("  %I64d\n",ans);
 91 
 92         //printf("%I64d\n",ans);
 93 
 94         for(j=4;j*j*j<=n;j++){
 95             ll te=n;
 96             int flag=1;
 97                 while(te){
 98                     ll yu=te%j;
 99                     if(yu!=3 && yu!=4 && yu!=5 && yu!=6){
100                         flag=0;break;
101                     }
102                     te/=j;
103                 }
104             if(flag==1) ans++;
105         }
106         printf("%I64d\n",ans);
107        // }
108 
109 
110     }
111     return 0;
112 }