hdoj 4937 Lucky Number 【思维题】

Lucky Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1487    Accepted Submission(s): 441

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.
 

 

定义：若一个数n的k进制全部由3、4、5、6这些数组成，称k是n的幸运数。

题意：给你一个数n(1 <= n <= 1e12)，问你它的幸运数有多少个。若不存在输出0，无穷多输出-1。

思路：n最多为1e12，就算一层for也会TLE。我们可以考虑把搜索范围尽可能的缩小。

首先，我们可以把n写成n = a0 + a1 * k^1 + a2 * k^2 + ...

情况一：只有a0组成，直接判断即可。

情况二：n = a0 + a1 * k^1，这时直接求解就可以了，k的范围 >= 1e7

情况三：n = a0 + a1 * k^1 + a2 * k^2，这时就是求解一元二次方程了，k的范围 >= 1e5

其余情况：会出现k^3，这时搜索范围k <= (n)^(1/3)，时间复杂度可以承受，直接暴力即可。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#define LL long longusing namespace std;int kcase = 1;bool judge(LL base, LL n){    while(n)    {        if(n % base < 3 || n % base > 6)            return false;        n /= base;    }    return true;}void solve(){    LL n;    scanf("%lld", &n);    printf("Case #%d: ", kcase++);    if(n >= 3 && n <= 6)    {        printf("-1\n");        return ;    }    LL ans = 0;    for(LL a = 3; a <= 6; a++)    {        for(LL b = 3; b <= 6; b++)        {            if((n-a) % b) continue;            if((n-a) / b > max(a, b))                ans++;        }    }    for(LL c = 3; c <= 6; c++)    {        for(LL b = 3; b <= 6; b++)        {            for(LL a = 3; a <= 6; a++)            {                LL C = c - n;                double temp = sqrt(b*b - 4*a*C);                if(temp == (LL)temp)                {                    LL d = (LL)temp;                    if((d-b) % (2*a) == 0)                    {                        LL x = (d-b) / (2*a);                        if(x > max(max(a, b), c))                            ans++;                    }                }            }        }    }    for(LL base = 2; base * base * base <= n; base++)        if(judge(base, n))            ans++;    printf("%lld\n", ans);}int main(){    int t;    scanf("%d", &t);    while(t--){        solve();    }    return 0;}