HDU 1018 Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 34053    Accepted Submission(s): 16100

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

21020

 

Sample Output

719

题目分析：大体上就是先输入一个整数N，然后分别再输入N个整数，分别判断每一个整数的阶乘有多少位，比如输入10,10！=3628800，这个数一共有7位，因此输出7；

当然，这道题目肯定不能直接求出数的阶乘再判断位数，这样会溢出的，下面我用了两种方法来求，都实现了AC，先介绍第一种：

采用斯特林公式求解：

（斯特林公式：lnN!=NlnN-N+0.5*ln(2*N*PI)）求出位数即可，还是要注意一下输出格式哦！贴代码：

#include<iostream>#include<math.h>using namespace std;#define PI 3.1415926535int main(){    int T;//记录每一块    int result;  //记录下结果     cin>>T;    while(T--)    {        int n;        double t;        cin>>n;        t=(n*log(n)-n+0.5*log(2*n*PI))/log(10);        result=(int)t+1;        cout<<result<<endl;      }      return 0; } 

当然了，用这种方法求解的前提要先知道这个数学公式，老实说我也是百度之后才知道的，不知道的人拿到手肯定无法用这种方法来做了，那就用第二种方法，这种方法学过对数的童鞋都应该会吧0.0！

lnN!=lnN+ln(N-1)+ln(N-2)+ln(N-3)......+ln2+ln1;

用循环求得1~N的对数和lnN!,然后[lnN!]，取对数和的上整数就可以了，这种方法也很简单。贴AC代码：

#include<iostream>#include<math.h>using namespace std;int main(){int T;//记录每一块cin>>T;while(T--){int n;double t=0.0;cin>>n;for(int i=1;i<=n;i++){t+=log(i)/log(10);}        cout<<(int)t+1<<endl; }  return 0; }