LightOJ1045 Digits of Factorial 求n的阶乘在k进制下的位数

Description
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
5
5 10
8 10
22 3
1000000 2
0 100
Sample Output
Case 1: 3
Case 2: 5
Case 3: 45
Case 4: 18488885
Case 5: 1

用换底公式去做即可，这样也能方便预处理节省时间。

logk( n! ) = log( n! ) / log( k )

#include <iostream>#include <cstdio>#include <map>#include <set>#include <vector>#include <queue>#include <stack>#include <cmath>#include <algorithm>#include <cstring>#include <string>using namespace std;#define INF 0x3f3f3f3ftypedef long long LL;double a[1000005];void p(){    int i=0;    a[0]=log(1);    for(int i=1;i<1000005;i++){        a[i]=a[i-1]+log(i);    }}int main(){std :: ios :: sync_with_stdio (false);    int t,n,k;    p();    scanf("%d",&t);    for(int i=1;i<=t;i++){        scanf("%d%d",&n,&k);        double ans=a[n]/log(k)+1;        printf("Case %d: %d\n",i,(int)ans);    }    return 0;}