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 |
#include<stdio.h>
#include<cmath>
using namespace std;
double a[1000010];
int main()
{
freopen("in.txt","r",stdin);
a[1]=log(1);
for(int i=2;i<=1000000;i++)a[i]=a[i-1]+log(i);
int t;
scanf("%d",&t);
for(int ca=1;ca<=t;ca++)
{
int n,b;
scanf("%d%d",&n,&b);
if(n==0)
{
printf("Case %d: %d\n",ca,1);
continue;
}
double ans=0;
printf("Case %d: %.0lf\n",ca,floor(a[n]/log(b*1.0))+1);
}
return 0;
}