Number Theory(数论-对数)
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Description
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) andbase (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
5
5 10
8 10
22 3
1000000 2
0 100
Sample Output
Case 1: 3
Case 2: 5
Case 3: 45
Case 4: 18488885
Case 5: 1
意解: 题意给出n和k,叫你求出n的阶乘的k进制数有多少位; 此处用到了对数的知识.
先说下对数的换底公式, 有loga ^ b = logc ^ b / log c ^ a; 容易证明这是正确的,高中的知识....
之后对于求一个数x的位数,我们先从10进制数说起,假设求100000的位数,我们知道其有6为数,
可以把其化为10^5,而易知log10(10^ 5) = 5,即其位数为log10(x) + 1;类比其他进制数就很自然了;
AC代码:
#include <iostream>#include <cstdio>#include <cmath>using namespace std;typedef long long ll;const int M = 1e6 + 100;double lo[M];void unit(){ for(int i = 1; i < M; i++) lo[i] = lo[i - 1] + log10(i);}int main(){ int T,n,k,cnt = 0; scanf("%d",&T); unit(); while(T--) { scanf("%d %d",&n,&k); printf("Case %d: %d\n",++cnt,(int)(lo[n] / log10(k)) + 1); } return 0;}
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