LightOJ - 1045 Digits of Factorial （数学公式） 对数换低公式

　　LightOJ - 1045

　　Digits of Factorial

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　　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 integerT (≤ 50000), denoting the number of test cases.

　　Each case begins with two integersn (0 ≤ n ≤ 106)andbase (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

　　Source

　　Problem Setter: Jane Alam Jan

　　//题意:输入n，m

　　给你一个数n，让你求n的阶乘数转换成m进制数的位数。

　　//思路：

　　用对数代替低公式loggg(a)(b)=log(c)(b)/log(c)(a);#include#include#include#include#include#define N 1000010using namespace std;double sum[N];int init(){for(int i=1;i<=N;i++)sum[i]=sum[i-1]+log10(i);}int main(){int t,T=1,n,m;init();scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);printf("Case %d: %d\n",T++,(int)(sum[n]/log10(m))+1);}return 0;}

　　Time Limit:2000MS

　　Memory Limit:32768KB

　　64bit IO Format:%lld & %llu