HDU 1018 Big Number

最新推荐文章于 2020-07-04 15:32:33 发布

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#integer #encryption #output #numbers #input #each

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本文介绍了一种高效计算大整数阶乘位数的方法，通过避免直接计算阶乘值来解决内存溢出问题。利用对数性质，文章提供了一个基于斯特林公式的实现方案，并附带了完整的C语言代码。

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16750 Accepted Submission(s): 7451

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

Source

Asia 2002, Dhaka (Bengal)

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JGShining

分析：题意就是求n的阶乘后的数字有几位。因为数据量很大所以不可能把阶乘算出来，所以构造令N！=10^t，然后两边取对数t=log10(N)+log(N-1)+……+log10(1).

floor(t)就是这个数字除最高位外的位数，然后再加上1（最高位还有1位）。数学公式（斯特林公式：lnN!=NlnN-N+0.5*ln(2*N*pi)）

代码：

#include<stdio.h>
#include<math.h>

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