light oj 1045 - Digits of Factorial(求阶乘在不同进制下的位数)

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：求阶乘在不同进制下的位数。

用对数搞一搞就好啦，设阶乘n在k进制下位数为sum，sum=(int)logk(n!)+1=(int)（log10(n!)/log10(k)）+1；然后以10为底打个表就好了。

注意n==0的时候特判一下。

#include
#include<string.h>
#include
#include
#define LL long long
using namespace std;
double sum[1000005];
void inin()
{
    sum[0]=0.0;
    for(int i=1; i<=1000000; i++)
        sum[i]=sum[i-1]+log10(i);
}
int main()
{
    inin();
    int T, t=1, n, k;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &k);
        if(n==0)printf("Case %d: 1\n", t++);
        else
        printf("Case %d: %d\n", t++, (int)(sum[n]/log10(k))+1);
    }
    return 0;
}