题目意思:
给你一个数,然后转化成相应进制的数,算出阶乘以后,求阶乘的位数
阶乘的位数我们这么来算:
例如1000的阶乘log10(1) + log10(2) + ...+log10(1000) 取整后加1
然后转化成进制的话就是: 除以log10(base) 后加1
题目:
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) 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
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
N!在十进制下的位数就是 log10 ( N ! ) +1 ( 自己找个数验证 ) ,N!在K进制下的位数就是 logK( N ! ) +1 ( K 为底数 )
而计算机不能直接以K为底求对数。所以运用换底公式, logK( N ! ) = log( N! ) / log ( K ) ,注意,等号右边的 log 都是
默认以e为底。
正式进入解题了,题目给出N,K,如果每一次输入N,K 都要计算 N!的话,那就有很大的开销,开销为O(N)。
所以可以采取一种办法,先预处理,用double数组 sum[ 1000000 ] 把 log ( N ! ) 存起来。数组 应该够的,需要说的是定义大
数组尽量放到外面,这样比较妥当,要不运行时会出错。
用sum [ i ] 表示 log(1 )+log(2)+。。。+log(i) , 这样时间开始时花费O(N),之后每次花费O(1),等输入的
时候用换底公式处理一下就得到答案了。
代码:
#include <cstdio>
#include <cstring>
#include <cmath>
double SumAdd[1000000 + 10];
int main(){
int cas = 0 , t , n , base , i ;
scanf("%d",&t);
for(i = 1 ; i < 1000010 ; i ++)
SumAdd[i] = SumAdd[i-1] + log(i * 1.0);
while(t--){
scanf("%d%d",&n,&base);
printf("Case %d: %d
",++cas,(int)(SumAdd[n] / log(base * 1.0))+1);
}
}