lightoj-1045

1045 - Digits of Factorial

PDF (English) Statistics Forum
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
3 5
4 4

1 1
10
1 1
Case 1:
1
3
12
Case 2:
10

解题思路: 因为求位数的时候可以用对数来求，所以f[]直接求对数存起来。 

位数=(f[n]/log(base)+1) (ps: +1 是因为例如 log2(4) = 2, 但他的位数是3来的, 所以要+1)

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;

typedef unsigned long long ull;

const int N = 1000010;
double f[N];

void init(){
    f[0] = log(1);
    for(int i=1;i<=N;i++){
        f[i] = f[i-1]+log(i*1.0); 
    }
    return ;
} 


int main(){
    init();
    int T,n,base;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        
        scanf("%d%d",&n,&base);
        
        printf("Case %d: %d
",t,(int)(f[n]/log(base)+1));
        
    }
    return 0;
}