杭电ACM OJ 1018 Big Number 两数相乘出一个大数，求大数的位数 注意log的使用

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39368    Accepted Submission(s): 19166

Problem Description

In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

 

Input

Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 10⁷ on each line.

 

Output

The output contains the number of digits in the factorial of the integers appearing in the input.

 

Sample Input

21020

 

Sample Output

719
翻译：输入一个数，比如10，求10的阶乘（10*9*8.。。*1）的位数
思路：因为仅仅求出位数即可，不用创造一套新的大数乘法运算，用log即可轻松求出位数。
1.为什么log可以求出位数 比如log10（100）那就是2，所以是2+1等于3，所以三位数，1000,10的3次，3+1就是4位数。
2.这题怎么应用log，首先你肯定会采用for循环，那么求位数肯定就是
log10（1*2*。。。n）=log10（1）+log10（2）。。。log10（n）
这就好了啊。
部分代码
private int getLength() {    double sum = 0;    for (int i = 1; i <= n; i ++) {        sum += Math.log10((double)i);    }    return (int)sum + 1;}
全部代码
public class BigNumber1018 {    private int n;    BigNumber1018(int n) {        this.n = n;    }    private int getLength() {        double sum = 0;        for (int i = 1; i <= n; i ++) {            sum += Math.log10((double)i);        }        return (int)sum + 1;    }    public static void main(final String[] args) throws Exception {        BigNumber1018 bigNumber1018 = new BigNumber1018(20);        int length = bigNumber1018.getLength();        System.out.println(length + "");    }}