Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

2
10
20

Sample Output

7
19

求大数阶乘的位数。。用java暴力写的。。后来一想我真是脑残10^7不T才怪，看到公式才觉得数学果然强大。

推导过程：
在这之前，我们必须要知道一个知识，任意一个正整数a的位数
等于(int)log10(a) + 1；为什么呢？下面给大家推导一下：

  对于任意一个给定的正整数a，
  假设10^(x-1)<=a<10^x，那么显然a的位数为x位，
  又因为
  log10(10^(x-1))<=log10(a)<(log10(10^x))
  即x-1<=log10(a)<x
  则(int)log10(a)=x-1,
  即(int)log10(a)+1=x
  即a的位数是(int)log10(a)+1

我们知道了一个正整数a的位数等于(int)log10(a) + 1，
现在来求n的阶乘的位数：
假设A=n!=1*2*3*......*n，那么我们要求的就是
(int)log10(A)+1，而：
	log10(A)
        =log10(1*2*3*......n)  （根据log10(a*b) = log10(a) + log10(b)有）
         =log10(1)+log10(2)+log10(3)+......+log10(n)
现在我们终于找到方法，问题解决了，我们将求n的阶乘的位
数分解成了求n个数对10取对数的和，并且对于其中任意一个数，
都在正常的数字范围之类。

总结一下：n的阶乘的位数等于
		  (int)(log10(1)+log10(2)+log10(3)+......+log10(n)) + 1


#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define ll long long
using namespace std;
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
	{
		scanf("%d",&n);
	    double ans=0;
	    for(int i=2;i<=n;i++)
			ans+=log10(i);
		printf("%d\n",1+(int)ans);
	}
    return 0;
}