杭电oj HDOJ 1018 Big Number

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1018

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.

题目大意

求每一个测试用例的阶乘的位数，测试用例的范围在1~10⁷

解题思路

由于本题测试用例的范围较大，如果使用“递归”和“循环”来解题一定会超时！所以要寻找另一种的解题方法。

计算一个数的位数：lg(n)+1lg(n)+1lg(n)+1因为本题中的数是10进制，所以对数是以10为底的，推广开来，m进制数n的位数为logmn+1{log}_{m}n+1logmn+1
计算一个数的阶乘：在这里使用“斯特林公式”n!≈2πn(ne)nn! \approx \sqrt[]{2\pi n}(\frac{n}{e})^nn!≈2πn(en)n然后两边取对数得：lg(n!)=lg(2πn)2+nlg(ne)lg(n!)=\frac{lg(2\pi n)}{2}+nlg(\frac{n}{e})lg(n!)=2lg(2πn)+nlg(en)
最后再把上面得到的对数加111就得出了n!n!n!的位数！

本人的C++解决方案

#define _USE_MATH_DEFINES
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;int main()
{int n, i, num, res;double temp1, temp2;cin >> n;for (i = 0; i < n; i++) {cin >> num;temp1 = log10(2 * M_PI * num);temp2 = log10(double(num) / M_E);res = int(temp1 / 2 + num * temp2);printf("%d\n", res + 1);}return 0;
}

代码通过HDOJ平台运行检查，如发现错误，欢迎指出和纠正，谢谢！