light-1045 - Digits of Factorial【思维】（数学）

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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

题意：求 n 的阶乘转化为 base 进制有多少位（求长度）？

题解：规律： log10 ( n ) = a + b （a是整数，b是小于1的小数），则 n 在 10 进制下的长度为 a + 1；那么推理可得 log（base）（n）= a + b，则在 base 进制下的长度是 a+1 位，但是计算机只能算 e 和 10 的对数，这就要用到高中数学的换底公式了；还需要注意的就是预处理一下。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
const int MAXN=1e6+10;
int n,base;
double a[MAXN];
void get()
{a[0]=0;for(int i=1;i<=MAXN;i++)a[i]=a[i-1]+log(i*1.0); // 这里利用了对数运算的性质把阶乘拆开了 log(n)(a*b) = log(n)(a)+log(n)(b); 
}
int main()
{int t,text=0;scanf("%d",&t);get();while(t--){scanf("%d %d",&n,&base);LL ans=1LL*(a[n]/log(base*1.0)+1);printf("Case %d: %lld\n",++text,ans);}return 0;
}

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