Lucky Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 294 Accepted Submission(s): 49
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
Mean:
给你一个10进制数n，现在要你找一个x，使得这个十进制数在x进制表示下的数字中只包含3,4,5,6这四个数字，问你这样的x有多少个。
analyse:
我们将n这个数在x进制下的表示记为：n=a0+a1*x+a2*x^2+a3*x^3+.....
我们采取枚举a0、a1、a2...，然后判断这个式子是否等于n的做法。讨论一下几种情况：1)a0：即a0==n，只有当a0等于3,4,5,6中其中一个的时候，才可能满足要求，而这种情况下，x取任何值都满足要求(当然最基本的条件x>a0要满足)，所以该种情况下就输出-1;2）a0+a1*x：此时要枚举的量有a0和a1，我们枚举在3,4,5,6中枚举a0和a1，那么如果方程：(n-a0)%a1==0成立(上面的基本条件不再重复)，此时也是成立的；3)a0+a1*x+a2*x^2：此时相当于求解方程a0+a1*x+a2*x^2=n这样的2次方程，但是要怎么解呢？利用求根公式：x=(-b±根号(b^2-4ac))/2a，然后判断这个值是否为整数就可以了。这样一来三位以内的x就被我们用枚举a0,a1,a2,a3的方式来枚举完了。我们可以证明a0+a1*x+a2*x^2+a3*x^3是可以将1e12以内的数表示出来的，以上三个步骤枚举完后，剩下的就是a*x^3这种情况，然后在x^3<n的范围内枚举进制就可以了、枚举x进制的就可以了。
Time complexity：不会超过O(n)开3次方次
Source code：
//Memory Time// 1347K 0MS// by : Snarl_jsb#include#include #include #include #include #include #include #include #include #include #include #define MAX 1100#define LL __int64using namespace std;int main(){ #ifndef ONLINE_JUDGE freopen("cin.txt","r",stdin); #endif int T,kase=0; cin>>T; while(T--){ LL n,t,ans=0; LL i,j,k; LL a,b,c,d,x; scanf("%I64d",&n); if(n>=3&&n<=6){ printf("Case #%d: -1\n",++kase); 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++; } } 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; d=(LL)sqrt(b*b-a*c*4+0.5); if(d*d!=b*b-a*c*4)continue; if((d-b)%(a*2))continue; x=(d-b)/(a*2); if(x>max(max(i,j),k)){ ans++; } } } } for(i=2;i*i*i<=n;i++){ t=n; while(t){ if(t%i<3||t%i>6) break; t=t/i; } if(!t){ ans++; } } printf("Case #%d: %I64d\n",++kase,ans); } return 0;}