LightOJ 1045 Digits of Factorial

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：求n！在base进制下的位数；

如果n是整数，log（n）+1表示n在十进制下的位数，log base （n ）+ 1 以base为底表示n在base进制下的位数，根据换底公式log base（n） =  log（n ）/ log （base），

所以log base （n！） = （ log（n）+ log（n-1） + log（n-2）…… log（1 ）） /  log（m）；先求出log前n项的和。

#include<cstdio>
#include<cmath>
#define M 1000010
#define ll long long
double sum[M];
int main()
{
ll t,i,k=0;
sum[0] = 0;
for(i = 1 ; i <= M ; i++)
{
sum[i] = sum[i-1]+log((double) i); //将i定义double型返回double值
}
scanf("%lld",&t);
while(t--)
{
ll n,m;
scanf("%lld %lld",&n,&m);
printf("Case %lld: ",++k);
if(n == 0)
{
printf("1\n");
continue;
}
ll ans = (ll)(sum
/log(m))+1; //ans输出整数
printf("%lld\n",ans);
}
}