Lightoj1045——Digits of Factorial（k进制的n的阶乘位数）

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

结论：k进制下的n的阶乘的位数=logk(n)=log10(n)/(log10(k))，所以可以先求出10进制下f
的对数打表。另外f
应先保持浮点型，最后向上取整

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<cmath>
#define MAXN 1000010
using namespace std;
double a[MAXN];
void Init()
{
a[0]=log10(1);
for(int i=1;i<=1000000;++i)
a[i]=a[i-1]+log10(i);
}
int main()
{
Init();
int ans;
int t,cnt=1,i,j,m,n;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
if(n==0)
ans=1;
else
{
ans=ceil(a
/log10(m));

}
printf("Case %d: %d\n",cnt++,ans);
}
return 0;
}