【LightOJ 1045 Digits of Factorial】

Digits of Factorial

actorial 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：计算N!在K进制下的位数,即计算log(1)+log(2)+…+log(N)其中log的底数都是K

思路：首先利用log(xy)=log(x)+log(y)可以求出log(n!)..然后换底公式求出log(b,n!)…然后加1即可。
询问次数很多，每次都重新算TLE的。
所以data[i]=1..i的log(i)之和,也就是前缀和的优化.然后O(1)回答每次询问。

#include<cstdio>#include<cmath>const int INF=1000011;double pa[INF];int main(){    int i;    pa[0]=0;    for(i=1;i<=INF;i++)        pa[i]=pa[i-1]+log(i);    int T,nl=1,N,M;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&N,&M);        printf("Case %d: %d\n",nl++,(int)(pa[N]/log(M))+1);    }    return 0;}