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枚举 + 进制转换 --- hdu 4937 Lucky Number

2024-07-17 05:49:25   217人阅读

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
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
 

 

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<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<string>#include<climits>#include<cmath>#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;}

　　


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