HDU 1018 Big Number 数学题

本文介绍了一种计算给定整数阶乘位数的方法。利用对数性质简化问题,通过累加每个整数的对数值来高效求解。提供了一个C++实现示例。
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
 
题目意思就是给你一个数,求该数的阶乘有多少位。
这里需要用到一个对数的知识
1.  一个数对于10的对数就是该数的位数
2.  log10(x*y)=log(x)+log(y)
知道上面的两条后,题目就相当简单了。
代码如下
 1 #include<cstdio>
 2 #include<cmath>
 3 #include<cstring>
 4 #include<iostream>
 5 using namespace std;
 6 int n;
 7 int main(){
 8     scanf("%d",&n);
 9     while(n--)
10     {   
11         int t;
12         scanf("%d",&t);
13         int ans=0;
14         double k=0;
15         for (int i=1;i<=t;++i)
16         {
17             k+=log10(i);
18         }
19         ans=(int)k+1;
20         printf("%d\n",ans);
21     }
22     return 0;
23 }

 

转载于:https://www.cnblogs.com/fakerv587/p/5179229.html

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