HDU 1018 Big Number数论（解题报告）

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

额。。。。一道数学题，还是很有意思的，受益了。

两种解法。

解法一：

   首先看一个数：1234567 很明显它是7位吧，然后我们会发现，10^6<1234567<10^7;

所以咱们就想到了对数了们可以得出 log10（1234567）=6.091514  设m = 6.091514 那么（int）m = 6 m+1就是答案了。

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

解法二：

   利用数学公式：斯特林公式：ln(N!) = Nln(N)-N + 0.5*ln(2*N*pi)

                                 还有对数换底公式啊：loge(N)/loge(10)=log10(N);

#include<iostream>#include<stdio.h>#include<math.h>#define pi 3.141592654using namespace std;int main(){int n,t;double sum;scanf("%d",&t);while(t--){scanf("%d",&n);double temp=(n*log(n)-n+0.5*log(2*n*pi))/log(10);printf("%d\n",(int)temp+1);}return 0;}