HDU 1018 大数问题

发布时间：2020-12-14 03:51:21 所属栏目：大数据来源：网络整理

导读：Big Number Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22912????Accepted Submission(s): 10325 Problem Description In many applications very large integers numbers are required. S

Big Number

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22912????Accepted Submission(s): 10325

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

题目大意：

? ? ? ? ? ?求一个数的阶乘最终输出结果的位数

解题方法：

? ?利用公式：

两种做法：

一、

N！=1*2*3....*n

求以10 为底的对数就可以了

log10(n!)=log10(1)+ log10(2) +log10(3)...+log10(n);

不过最终误差为 1 log10（10） = 1 位数相差1?

二、

斯特林数，第一类斯特林数就可以做这个！

斯特林数能够做一切关于阶乘有关的大数运算 ?具体内容自行百度

这里给出递归公式：

log10(n!)=1.0/2*log10(2*pi*n)+n*log10(n/e)

然后附上代码了;

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

int main()
{
    int n,i,t;
    cin>>n;
    while(n--)
    {
        cin>>t;
        double  ans = 1;
        for(i = 1; i <= t; i++)
        {
            ans += log10((double)i);
        }
        cout<<(int)ans<<endl;
    }
    return 0;
}

其次利用斯特林公式求解：

#include<iostream>
#include<cmath>

using namespace std;
#define pi 3.1415926
#define e  2.7182818284590452
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n ;
        cin>>n;
        int ans ;
        ans = 1.0/2*log10(2*pi*n) +n*log10(n/e);
        cout<<ans+1<<endl;

    }
    return 0;
}

（编辑：李大同）

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