乘法与位数

题目意思：

给你一个数，然后转化成相应进制的数，算出阶乘以后，求阶乘的位数

阶乘的位数我们这么来算：

例如1000的阶乘log10（1） + log10（2） + ...+log10（1000） 取整后加1

然后转化成进制的话就是： 除以log10（base） 后加1

题目：

A - Digits of Factorial

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

Submit  Status

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

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

#include <stdio.h>
#include <iostream>
#include <math.h>

using namespace std;

double a[1000005] = {0};
void cal () {
	for (int i = 1; i <= 1000005; i ++) {
		a[i] = a[i - 1] + log10(1.0 * i);
	}
}
int main () {
	int cas;
	int n, base;
	scanf("%d", &cas);
	cal();
	for (int t = 1; t <= cas; t ++) {
		scanf("%d%d", &n, &base);
		printf("Case %d: %d\n", t, (int)(a[n] / log10(base * 1.0) + 1));
	}
	return 0;
}