【Ligth-oj】-1045 - Digits of Factorial（数论，log，好）

1045 - Digits of Factorial

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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：就是给你一个数n，求出他的阶乘，把阶乘转换成k进制会有多少位。如5！=120,120化为8进制就是170,有3位

题解：把一个十进制数怎么换成k进制呢？就是每次除以k得到的余数就是一位。如：120换成8进制。120%8=0,120/8=15;

15%8=7,15/8=1;

1%8=1,1/8=0;那么就是170了(倒着写)

n!<=k^m,求出最小的m就是换算出k进制有多少位了。

1*2*384*......*n=k^m两边分别取对数得：log1+log2+log3+......+logn=m*logk;

那么m=(log1+log2+log3+......+logn)/(logk)

前面log1+....logn可以先打出来

AC代码：

<span style="font-size:10px;">#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define LL long long
double a[1000010];
void init()
{
a[1]=0;
for(int i=2;i<=1000000;i++)
a[i]=a[i-1]+log((double)i);		//求出 n!取对数的值
}
int main()
{
int u,ca=1;
init();
scanf("%d",&u);
while(u--)
{
int n,k;
scanf("%d%d",&n,&k);
printf("Case %d: ",ca++);
if(n==0)
printf("1\n");
else
{
double ans;
ans=a
/(log((double)k));
if(ans!=(int)ans)
ans=(int)ans+1;
printf("%.lf\n",ans);
}
}
return 0;
}</span><span style="font-size:18px;">
</span>