LightOJ 1045 Digits of Factorial（大数进制）

http://lightoj.com/volume_showproblem.php?problem=1045

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

 

PROBLEM SETTER: JANE ALAM JAN

题意：

求解 n! 在 base 进制下有几位。

如 5！=120 ，10进制下是 3 位。

思路：

大数的进制转换，也是学习了。

首先应理解的十进制化为二进制时使用的方法“取模法”，这里参考的博客：博客园

另外，有一个公式可以学下：

参考代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MYDD=1103+1e6;

double s[MYDD];
void Init() {/*初始化*/
s[0]=s[1]=0;
for(int j=2; j<MYDD; j++) {
s[j]=s[j-1]+log(j*1.0);
}
}

int main() {
int tt,kc=1;
scanf("%d",&tt);
Init();
while(tt--) {
int n,b;
scanf("%d%d",&n,&b);
if(n==0) {
printf("Case %d: 1\n",kc++);
continue;
}
double ans=s
;
ans=ans/log(b*1.0);
printf("Case %d: %d\n",kc++,(int)ans+1);
}
return 0;
}