【LightOJ 1045 Digits of Factorial】

Digits of Factorial

actorial 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：计算N!在K进制下的位数,即计算log(1)+log(2)+…+log(N)其中log的底数都是K

思路：首先利用log(xy)=log(x)+log(y)可以求出log(n!)..然后换底公式求出log(b,n!)…然后加1即可。

询问次数很多，每次都重新算TLE的。

所以data[i]=1..i的log(i)之和,也就是前缀和的优化.然后O(1)回答每次询问。

#include<cstdio>
#include<cmath>
const int INF=1000011;
double pa[INF];
int main()
{
int i;
pa[0]=0;
for(i=1;i<=INF;i++)
pa[i]=pa[i-1]+log(i);
int T,nl=1,N,M;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&N,&M);
printf("Case %d: %d\n",nl++,(int)(pa
/log(M))+1);
}
return 0;
}