【lightoj1045】数学-小知识点

最新推荐文章于 2019-03-19 22:16:49 发布

原创最新推荐文章于 2019-03-19 22:16:49 发布 · 401 阅读

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本文介绍了一种高效算法，用于计算整数n的阶乘在特定进制k下的位数。通过预先计算和存储中间结果，该算法能够在短时间内得出答案，适用于解决诸如竞赛编程中涉及的大数问题。

E - E 使用long long

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1045

uDebug

Description

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

5 10

8 10

22 3

1000000 2

0 100

Sample Output

Case 1: 3

Case 2: 5

Case 3: 45

Case 4: 18488885

Case 5: 1

有个结论：以k进制求n！结果的位数为logk（n!）+1，因为计算机只能进行以e，10，也有2为底的阶乘，所以计算前要用换底公式处理一下，这个以前都学过logk(n!)=log(n!)/log(k)其中等号后面都是默认以e为底的。要节省时间先打表预处理一下。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 1e6+10;
double a[N];
int main() {
	int T,p=0;
	scanf("%d",&T);
	for(int i=1; i<=N; i++) {
		a[i]=a[i-1]+log(i);
	}
	while(T--) {
		int n,base;
		double ans;
		scanf("%d%d",&n,&base);
		if(n==0)
			printf("Case %d: 1\n",++p);
		else {
			ans=a[n]/log(base)+1;
			printf("Case %d: %d\n",++p,(int)ans);
		}
	}
	return 0;
}