light oj 1045 - Digits of Factorial K进制下N！的位数
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 |
分析：换底公式log a b = log c b / log c a; 所以logk(fn) = log10(fn)/ log10k; logq0(fn) = log10(N) = log10(1 * 2 *...*n) = log10(1) + 1og10(2) .....+ 1og10(n)
代码：
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define N 1000009
using namespace std;
double a[N];
void init()
{
a[0] = log10(1);
for(int i = 1; i <= N; i++)
a[i] = a[i-1] + log10(i*1.0);
}
int main(void)
{
int T, cas;
int n, k;
init();
scanf("%d", &T);
cas = 0;
while(T--)
{
cas++;
scanf("%d%d", &n, &k);
if(n == 0)
printf("Case %d: 1\n", cas);
else
{
double ans = ceil(a[n]/log10(k*1.0));
printf("Case %d: %d\n", cas, (int)ans);
}
}
