Big Number（简单数学题）

最新推荐文章于 2024-11-13 14:47:51 发布

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最新推荐文章于 2024-11-13 14:47:51 发布

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本文介绍了一种通过对数来计算给定整数阶乘位数的方法，并提供了相应的AC代码实现。

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Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26003 Accepted Submission(s): 11810

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 ≤ 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

Sample Output

  
   7
19

Source

Asia 2002, Dhaka (Bengal)

意解：用对数来求n的阶乘的位数；
AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;
typedef long long ll;

int main()
{
    int n,m;
    cin>>m;
    while(m--)
    {
        scanf("%d",&n);
        double sum = 0;
        for(int i = 1; i <= n; i++)
        {
            sum += log10((double)i);
        }
        printf("%d\n",(int)sum + 1);
    }
    return 0;
}