HDU 1018-Big Number（数学）

Big Number

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

Total Submission(s): 26383    Accepted Submission(s): 12006

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。看到公式才认为数学果然强大。
推导过程：

在这之前，我们必需要知道一个知识，随意一个正整数a的位数
等于(int)log10(a) + 1。为什么呢？以下给大家推导一下：

  对于随意一个给定的正整数a。
  如果10^(x-1)<=a<10^x，那么显然a的位数为x位，
  又由于
  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，
如今来求n的阶乘的位数：
如果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)  （依据log10(a*b) = log10(a) + log10(b)有）
         =log10(1)+log10(2)+log10(3)+......+log10(n)
如今我们最终找到方法，问题攻克了。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，而且对于当中随意一个数，
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;
}

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