hdu 1018 Big Number

最新推荐文章于 2025-09-11 23:50:21 发布

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原文链接：http://www.cnblogs.com/pshw/p/4797638.html

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#java

本文探讨了如何通过数学公式计算一个数的阶乘所包含的位数，涉及对数运算和数学推理，特别适用于处理非常大的整数。

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31296 Accepted Submission(s): 14539

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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一道和数学有关的题，必须用到一个数学公式，一个数n的位数等于 log10（n）+1 。

题意：输入n，求n！是一个几位数。

附上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 using namespace std;
 5 
 6 int main()
 7 {
 8     int n,i,m;
 9     double s;
10     scanf("%d",&n);
11     while(n--)
12     {
13         s=0.0;
14         scanf("%d",&m);
15         for(i=1; i<=m; i++)
16             s+=(log10(i));
17         printf("%d\n",(int)s+1);
18     }
19     return 0;
20 
21 }