HDU 1018 Big Number（简单数学解法无需高精度）

• Big Number

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 ≤ 10 7 on each line.

• Output

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

• Sample Input

2
10
20

• Sample Output

7
19

题意：给出一个整数n，范围1-10^7，计算n的阶乘的位数。

思路一：一个数的可以由10^n表示，n+1代表数的位数，例如100是10^2，位数是2+1，在此基础上我们想要求出一个数的n，很容易想到用log以10为底取对数便可得到n，然后加一就得到了数的位数。那么x!的位数为log10(x!)+1，x! = x * (x-1) * ( x - 2) * ……*1，由此可得，只需要一层循环便可得到答案，AC代码如下：

#include 
#include 
using namespace std;

int main()
{
	int n;
	scanf("%d", &n);
	while(n--)
	{
		int x;
		double num = 0.0;
		scanf("%d", &x);
		for(int i = 1; i <= x; i++)
		{
			num += log10(i);
		}
		printf("%.0lf\n", num + 0.5);  //因为输出会四舍五入所以加0.5就可以了，如果num为7.5那么答案就为8，如果加1的话为8.5，输出就是9，就会导致答案出错
	}
	return 0;
} 

==思路二：==其实这种方法与上一种类似，，根据斯特林公式
根据此公式可直接得到n!，然后用上面的方法求出位数。
代码如下：

#include 
#include 
using namespace std;
#define PI acos(-1)
#define e 2.71828182845
int main()
{
	int n;
	scanf("%d", &n);
	while(n--)
	{
		int x;
		double num = 0.0;
		scanf("%d", &x);
		num = log10(sqrt(2 * PI *x)) + x * log10(x / e);
		printf("%.0lf\n", num + 0.5);
	}
	return 0;
} 

End