http://lightoj.com/volume_showproblem.php?problem=1045

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.

?

题意：

求解 n! 在 base 进制下有几位。

如 5！=120 ，10进制下是 3 位。

思路：

大数的进制转换，也是学习了。

首先应理解的十进制化为二进制时使用的方法“取模法”，这里参考的博客：博客园

另外，有一个公式可以学下：

参考代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MYDD=1103+1e6;

double s[MYDD];
void Init() {/*初始化*/
	s[0]=s[1]=0;
	for(int j=2; j<MYDD; j++) {
		s[j]=s[j-1]+log(j*1.0);
	}
}

int main() {
	int tt,kc=1;
	scanf("%d",&tt);
	Init();
	while(tt--) {
		int n,b;
		scanf("%d%d",&n,&b);
		if(n==0) {
			printf("Case %d: 1n",kc++);
			continue;
		}
		double ans=s[n];
		ans=ans/log(b*1.0);
		printf("Case %d: %dn",kc++,(int)ans+1);
	}
	return 0;
}

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