HDU-1018（n!的位数）

最新推荐文章于 2024-07-28 10:45:00 发布

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最新推荐文章于 2024-07-28 10:45:00 发布

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分类专栏：题解文章标签：数学水题

本文链接：https://blog.youkuaiyun.com/qq_42804678/article/details/84727759

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本文探讨了在处理大整数阶乘时如何高效计算其位数的方法，适用于安全传输数据、加密等应用中。通过使用对数性质，提供了一种计算n!位数的算法，并给出了具体的实现代码。

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原题传送

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44891 Accepted Submission(s): 21955

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

题意：求n!的位数

公式：n的阶乘的位数=（int）log10(n*(n-1)*(n-2).....*2*1)+1=（int）（log10（n）+……log10（1））+1；

AC代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
	int n,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		double sum=0;
		for(int i=1;i<=n;i++)	
		sum+=log10(i);
		printf("%d\n",(int)sum+1);//求位数(int)log10(x)+1 
	}
	return 0;
 }