LIGHTOJ 1045 - Digits of Factorial 【阶乘取对数】

1045 - Digits of Factorial 

Time Limit: 2 second(s) Memory Limit: 32 MB

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

题意：求出n!用base进制表示有多少位；

思路：n=10^a*10^b->log(10,n)=a+b ，a是整数表示n在10进制下10的最高次方，b是小数，10^b表示剩余的数的大小，a+1就是n在10进制下的长度了，在base进制下：log(base,N!)={ln(1)+ln(2)+ln(3)+....+ln(N)}/ln(base),  在处理时预先处理一下 a[n]表示ln(1)+..ln(n);

失误：没有想到还能这样写，看来对数和阶乘还是紧密行相连的，阶乘取对数就成累加了，还能表示一个数的最高次方；

代码如下：

#include<cstdio>#include<math.h>double a[1000000+99];void init()//区间值打个表表示 {int i=0; a[1]=0.0;for(i=1;i<=1000000+12;++i){double tem=i*1.0;a[i]=a[i-1]+log(tem);}}int main(){int T,N,base,Kase=0;init();scanf("%d",&T);while(T--){scanf("%d %d",&N,&base);int ans=(int)(a[N]/(a[base]-a[base-1]))+1;printf("Case %d: ",++Kase);printf("%d\n",ans);}return 0; }