lightoj 1045 - Digits of Factorial 【数学】

题目链接：lightoj 1045 - Digits of Factorial

1045 - Digits of Factorial

PDF (English) Statistics Forum

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

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：给定一个数n，用base进制表示n!。

思路：换底公式 搞搞就好了。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 1e6 +10;
const int INF = 0x3f3f3f3f;
double ans[MAXN];
int main()
{
ans[0] = 0;
for(int i = 1; i <= 1000000; i++) {
ans[i] = ans[i-1] + log10(i*1.0);
}
int t, kcase = 1; scanf("%d", &t);
while(t--) {
int n, base; scanf("%d%d", &n, &base);
if(n == 0) {
printf("Case %d: 1\n", kcase++);
continue;
}
printf("Case %d: %lld\n", kcase++, (LL)ceil(ans
/ log10(base*1.0)));
}
return 0;
}