Digits of Factorial (求N!的k进制位数)（打表+log数学公式）

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

题意：n！转化成k进制会有多少位。

题解：

先举一个例子：

5！=120,120化为8进制就是170,有3位。

我们先看十进制120如何转成八进制170的。

1.120%8=0,120/8=15

2.15%8=7，15/8=1

3.1%8=1,     1/8=0;

这样三次就出来170，且正好3位

即n!<=k^m,求出最小的m就是换算出k进制有多少位了。

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

那么m=(log1+log2+log3+......+logn)/(logk)

前面log1+....logn可以先打表出来

#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[0] = a[1] = 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;
}