lightoj1045 - Digits of Factorial

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

 

loga(b)即为b在a进制下的位数，lg(a*b)=lg(a)+lg(b);由此可求出N！在十进制下的位数然后利用换底公式即可求出在K进制下的位数

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>using namespace std;const int maxn=1000010;double F[maxn];void init(){F[0]=0;for(int i=1;i<maxn;++i){F[i]=log10(1.0*i)+F[i-1];}}int main(){init();int t,i,j,n,k,test=1;scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);if(n==0){printf("Case %d: 1\n",test++);}else printf("Case %d: %d\n",test++,(int)ceil(F[n]/log10(k*1.0)));}return 0;}