HDU 1018Big Number（大数的阶乘的位数，利用公式）

Posted 2020-12-04 yinbiao

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1018

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42715    Accepted Submission(s): 20844

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

 

 

Source

Asia 2002, Dhaka (Bengal)

 

分析：

问你一个数的阶乘是多少位

n比较大

是大数问题

一开始不知道写

看了一下大佬的博客

 

具体分析；

 一个正整数a的位数等于(int)log10(a) + 1
    假设A=n!=1*2*3*......*n，那么我们要求的就是
    (int)log10(A)+1，而：
    log10(A)
    =log10(1*2*3*......n)  （根据log10(a*b) = log10(a) + log10(b)有）
    =log10(1)+log10(2)+log10(3)+......+log10(n)

 

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define max_v 105
int main()
{
    /*
    一个正整数a的位数等于(int)log10(a) + 1
    假设A=n!=1*2*3*......*n，那么我们要求的就是
    (int)log10(A)+1，而：
    log10(A)
    =log10(1*2*3*......n)  （根据log10(a*b) = log10(a) + log10(b)有）
    =log10(1)+log10(2)+log10(3)+......+log10(n)
        */
    int n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        double sum=0;
        for(int i=1; i<=n; i++)
        {
            sum+=log10((double)i);
        }
        printf("%d
",(int)sum+1);
    }
    return 0;
}