【LightOJ】1045 - Digits of Factorial（数论）

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1045 - Digits of Factorial

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.

题目要求n的阶乘变成k进制后的位数。如5的阶乘变成8进制是170，三位数。

我们来考虑一下k进制是怎么由十进制转化得到的——对了，辗转相除法。每一次除以k的余数作为一位数。

那么我们可以列如下方程：

n! <= k^x

求出最小的x就是结果了。

我们还是需要化简：两边同时求log

得：logn! <= x*logk

把n的阶乘分开写：log1 + log2 + ... + logn <= x*logk

那么x就很明显了，做个除法向上取整就行了。

然后要注意直接for循环求会TLE，应该先打表预处理一下。

代码如下：

#include <stdio.h>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
#define PI acos(-1)
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define LL long long
double a[1000000+11];
void getlog()
{
	a[0] = a[1] = 0;
	for (int i = 2 ; i <= 1000000 ; i++)
		a[i] = a[i-1] + log((double)i);
}
int main()
{
	getlog();
	int n,k;
	int u;
	int Case = 1;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d %d",&n,&k);
		printf ("Case %d: ",Case++);
		if (n == 0)
		{
			puts("1");
			continue;
		}
		double ans = a[n];
		ans = ans / (log((double)k));
		if (ans != (int)ans)
			ans = (int)ans + 1;
		printf ("%.lf\n",ans);
	}
	return 0;
}