LightOJ 1045 I - Digits of Factorial

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

题意：n!的m进制一共多少位————————————-

想到了怎么预处理，可惜没想到对数这个工具orz

log(n!) = log(n)+log(n-1) + log(n-2) +…+log(2)

例子：4！=24,24是十进制的2位，因为>10,5!=120，是十进制的3位，因为>100,所以直接对数化

反思：当题目要求求位数的时候应该要想到对数这个工具。

log（i）为自然对数

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
const int N=1000005;
double num
;
int main (void)
{
int t,cas=0;
for(int i=1;i<=N-1;i++)
{
num[i]=num[i-1]+log(i);
}
cin>>t;
while(t--)
{
int n,base;
cas++;
scanf("%d %d",&n,&base);
int ans;
ans=floor(num
/log(base)+1);//floor向下取整
//事实证明不用floor也行
printf("Case %d: %d\n",cas,ans);
}
return 0;
}