标签：cert mst val number tin one limit strong diff
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 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.
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 |
分析：换底公式log a b = log c b / log c a; 所以logk(fn) = log10(fn)/ log10k; logq0(fn) = log10(N) = log10(1 * 2 *...*n) = log10(1) + 1og10(2) .....+ 1og10(n)
代码：
| 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 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.
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 |
light oj 1045 - Digits of Factorial K进制下N！的位数
标签：cert mst val number tin one limit strong diff
原文地址：http://www.cnblogs.com/dll6/p/7266815.html
