1045 - Digits of Factorial
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Time Limit: 2 second(s)    Memory Limit: 32 MB
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

5！=120,120化为8进制就是170,有3位。

1.120%8=0,120/8=15

2.15%8=7，15/8=1

3.1%8=1,     1/8=0;

1*2*3*......*n=k^m两边分别取对数得：log1+log2+log3+......+logn=m*logk;

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<set>#include<queue>#include<stack>#include<map> using namespace std;typedef long long ll;     double a[1000000+11];void getlog(){  a = a = 0;  for (int i = 2 ; i <= 1000000 ; i++)    a[i] = a[i-1] + log((double)i);}int main(){  getlog();  int n,k;  int T;  int cas = 1;  scanf ("%d",&T);  while (T--)  {    scanf ("%d %d",&n,&k);    printf ("Case %d: ",cas++);    if (n == 0)    {      printf ("1\n");      continue;    }    double ans = a[n];    ans = ans / (log((double)k));    if (ans != (int)ans)      ans = (int)ans + 1;    printf ("%.lf\n",ans);  }  return 0;}