lightoj 1045 - Digits of Factorial (数学－log运用--阶乘进制位数)

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：是求N！的k进制的位数

不会处理，看的题解，原来有特殊解题方法

思路：要知道一个求进制位数的公式：计算 [ log(1)+log(2)+...+log(N) ]+1  其中log的底数都是K



然后就是用公式求出就好了

代码：

#include<cstdio>
#include<cmath>
double sum[1000010];
int main()
{
int t,kcase=1;
int n,m;
sum[0]=0;
for(int i=1;i<1000010;i++)//t和n的范围都很大，所以先储存到1000010，然后直接使用 ;注意是double型
sum[i]=sum[i-1]+log(i);
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
if(n==0)
printf("Case %d: 1\n",kcase++);
else
{
double ans=sum
/log(m)+1; //换底公式
printf("Case %d: %lld\n",kcase++,(long long)ans);
}
}
return 0;
}