light oj 1045 - Digits of Factorial K进制下N！的位数

1045 - Digits of Factorial

Factorial of an integer is defined by the following function

f(0) = 1

f(n) = f(n - 1) * n, if(n > 0)

So, factorial of 5 is 120. But in different bases, the factorial may be different. For example, factorial of 5 in base 8 is 170.

In this problem, you have to find the number of digit(s) of the factorial of an integer in a certain base.

Input

Input starts with an integer T (≤ 50000), denoting the number of test cases.

Each case begins with two integers n (0 ≤ n ≤ 10⁶) and base (2 ≤ base ≤ 1000). Both of these integers will be given in decimal.

Output

For each case of input you have to print the case number and the digit(s) of factorial n in the given base.

Sample Input	Output for Sample Input
5 5 10 8 10 22 3 1000000 2 0 100	Case 1: 3 Case 2: 5 Case 3: 45 Case 4: 18488885 Case 5: 1

分析：换底公式log a b = log c b / log c a; 所以logk(fn) = log10(fn)/ log10k;    logq0(fn) = log10(N) = log10(1 * 2 *...*n) = log10(1) + 1og10(2) .....+ 1og10(n)

代码：

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define N 1000009

using namespace std;

double a[N];

void init()

{

a[0] = log10(1);

for(int i = 1; i <= N; i++)

a[i] = a[i-1] + log10(i*1.0);

}

int main(void)

{

int T, cas;

int n, k;

init();

scanf("%d", &T);

cas = 0;

while(T--)

{

cas++;

scanf("%d%d", &n, &k);

if(n == 0)

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

else

{

double ans = ceil(a[n]/log10(k*1.0));

printf("Case %d: %d\n", cas, (int)ans);

}

}