light oj1028 - Trailing Zeroes (I)

Posted 2020-09-30 缄默。

1028 - Trailing Zeroes (I)

 

We know what a base of a number is and what the properties are. For example, we use decimal number system, where the base is 10 and we use the symbols - {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. But in different bases we use different symbols. For example in binary number system we use only 0 and 1. Now in this problem, you are given an integer. You can convert it to any base you want to. But the condition is that if you convert it to any base then the number in that base should have at least one trailing zero that means a zero at the end.

For example, in decimal number system 2 doesn‘t have any trailing zero. But if we convert it to binary then 2 becomes (10)₂ and it contains a trailing zero. Now you are given this task. You have to find the number of bases where the given number contains at least one trailing zero. You can use any base from two to infinite.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer N (1 ≤ N ≤ 10¹²).

Output

For each case, print the case number and the number of possible bases where N contains at least one trailing zero.

Sample Input	Output for Sample Input
3 9 5 2	Case 1: 2 Case 2: 1 Case 3: 1

Note

For 9, the possible bases are: 3 and 9. Since in base 3; 9 is represented as 100, and in base 9; 9 is represented as 10. In both bases, 9 contains a trailing zero.

 

分析：题目大意是一个数转化成任意进制后末尾有0的种数 ，就是求一个数的因子数，转化成n进制后末尾有0的数一定是原十进制数能被n整除，也就是说如果一个数是n的倍数的时候，一定能转换成n进制数末尾时0.

素数分解的唯一性：p1^x1*p2^x2...pn^xn(一个整数可唯一地分解为一些不同质因子的若干次方的乘积)

再根据乘法原理 因子个数为(x1+1)*(x2+1) + ... + (xn + 1)

 

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>

using namespace std;
typedef long long ll;
const ll N=1e6+10;
ll vis[N], prime[N];
ll k;

void init()
{
k = 0;
memset(vis,0,sizeof(vis));

for(ll i = 2; i <= N; i++)
{
if(!vis[i])
{
prime[k++] = i;

for(ll j = i * i; j <= N; j += i)
vis[j] = 1;
}
}
}
int main()
{
int T, cas;
ll n, ans;

init();

scanf("%d", &T);
cas = 0;
while(T--)
{
cas++;
ans = 1;

scanf("%lld", &n);

for(ll i = 0; i < k && prime[i] * prime[i] <= n; i++)
{
if(prime[i] > n)
break;

if(n % prime[i] == 0)
{
long long sum = 1;

while(n % prime[i] == 0)
{
sum++;
n /= prime[i];

}
ans *= sum;
}
}

if(n > 1)
ans *= 2;

printf("Case %d: %lld\n", cas, ans-1);

}

return 0;
}