Light OJ：1045 Digits of Factorial（数学+思维+对数）

1045 - Digits of Factorial

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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

 

PROBLEM SETTER: JANE ALAM JAN

题目大意：问你n的阶乘在base进制下有多少位。
解题思路：所需要的知识点1：





所需要的知识点2：k进制下n的位数为（int）logk（n）+1.可以联想下十进制。
所需要的知识点3：换底公式，由于系统没有以任意数k为底计算对数的函数，那么我们用换底公式，logk(n!)=lg(n!)/lg(k)，lg代表以e为底，当然以10为底也行。
所需要的知识点4：lg(n!)=lg(1*2*3*4*5...*n)=lg(1)+lg(2)+lg(3)+....+lg(n)。
好了，代码如下：

#include <cstdio>
#include <cmath>
double sum[1000002];
int main()
{
sum[1]=0;
for(int i=2;i<1000002;i++)//给lg(n!)打表
{
sum[i]=sum[i-1]+log(i);//默认以e为底
}
int t;
scanf("%d",&t);
int kcase=1;
int n,k;
while(t--)
{
scanf("%d%d",&n,&k);
int ans=(int)(sum
/log(k))+1;//知识点2
printf("Case %d: %d\n",kcase++,ans);
}
return 0;
}