hdu-Big Number

最新推荐文章于 2020-07-24 11:37:03 发布

原创最新推荐文章于 2020-07-24 11:37:03 发布 · 8.7k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#jtahstu

hduoj 专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一种计算大整数阶乘位数的方法，通过使用对数运算来避免大整数运算的问题，提供了两种不同的实现方案，并附带样例输入输出。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Big Number

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

Total Submission(s): 26872 Accepted Submission(s): 12225

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)

Recommend

JGShining | We have carefully selected several similar problems for you: 1013 1017 1016 1002 1071

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;

int main()
{
    int t,n,i;
    double sum;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        sum=1;
        for(i=1;i<=n;i++)
            sum+=log10(i);
        cout<<(int)sum<<endl;

    }
    return 0;
}

</pre><pre class="cpp" name="code">#include <iostream>//斯特林公式求对数
#include <cmath>
using namespace std;
#define PI 3.141592654
int main()
{
    int n,t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        cout<<(int)((0.5*log(PI*2*n)+n*log(n)-n)/log(10))+1<<endl;
    }
    return 0;
}