Light-oj-1045 lightoj1045 - Digits of Factorial (N！不同进制的位数)

本文链接：https://blog.youkuaiyun.com/lu_1110/article/details/52294613

该博客介绍了如何计算一个整数阶乘在特定进制下的位数。通过给定的输入示例，解释了如何处理测试用例，并提供了计算阶乘位数的数学原理，包括换底公式和求和对数的方法。

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Factorial （阶层）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) andbase (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 10

8 10

22 3

1000000 2

0 100

Output for Sample Input

Case 1: 3

Case 2: 5

Case 3: 45

Case 4: 18488885

Case 5: 1

题意：N是十进制的数，求N！在K进制下的位数。

N！在十进制下的位数就是 log10 ( N ! ) +1 ( 自己找个数验证 ) ，N！在K进制下的位数就是 logK( N ! ) +1 ( K 为底数 )，而计算机不能直接以K为底求对数。所以运用换底公式， logK( N ! ) = log( N! ) / log ( K) ，注意，等号右边的 log 都是默认以e为底。

计算N！在K进制下的位数，即计算 [ log(1)+log(2)+...+log(N) ]+1 其中log的底数都是K。

#include <stdio.h>  
#include <math.h>  
#include <string.h>  
using namespace std;  
double sum[1000008];   
int main()  
{  
    double res;  
    int T, n, a, i;  
    int num = 1;  
    memset(sum, 0, sizeof(sum));  
    for(i = 1; i < 1000008; i++)  
        sum[i] =sum[i-1]+log(i);  
    scanf("%d", &T);  
    while(T--)  
    {  
        scanf("%d %d", &n, &a);  
        if(n == 0) printf("Case %d: 1\n", num++);  
        else  
        {  
            res = sum[n];  
            res = res/log(a)+1;  
            printf("Case %d: %d\n", num++, (int)res);  
        }  
    }  
    return 0;  
}