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
/log10(k*1.0)));
}
return 0;
}