Light OJ 1045 Digits of Factorial（求位数）

最新推荐文章于 2018-01-06 14:08:00 发布

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最新推荐文章于 2018-01-06 14:08:00 发布

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分类专栏： Math 文章标签：数学

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本文详细介绍了如何解决在特定基础上计算阶乘位数的问题，通过使用对数方法进行高效计算，并提供了关键代码实现。深入探讨了阶乘在不同基数下的变化规律及其在编程中的实际应用。

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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 10

8 10

22 3

1000000 2

0 100

Case 1: 3

Case 2: 5

Case 3: 45

Case 4: 18488885

Case 5: 1

与白书里面推荐的一道UVA题目很像， http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1002

也是利用对数来求位数。

代码：

#include<stdio.h>
#include<cmath>
using namespace std;
double a[1000010];
int main()
{
  freopen("in.txt","r",stdin);
  a[1]=log(1);
  for(int i=2;i<=1000000;i++)a[i]=a[i-1]+log(i);
  int t;
  scanf("%d",&t);
  for(int ca=1;ca<=t;ca++)
  {
    int n,b;
    scanf("%d%d",&n,&b);
    if(n==0)
    {
      printf("Case %d: %d\n",ca,1);
      continue;
    }
    double ans=0;
    printf("Case %d: %.0lf\n",ca,floor(a[n]/log(b*1.0))+1);
  }
  return 0;
}