航电1018 Big Number

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

2 10 20

Sample Output

7 19

Source

Asia 2002, Dhaka (Bengal)

 

Recommend

JGShining   |   We have carefully selected several similar problems for you:   1017  1016  1071  1019  1021 

 

 

/***********************************************************起初的代码*********************************************************************

 

#include<iostream>
#include<cmath>
using namespace std;

int get(int m){
	int pass = 0;        //进位 
	int num[100];        //存储位数 
	num[0] = 1;          //一个数的第几位 
	int count = 0;       //记录进位的位数 
	int s = 0;
	int number = 1;      //记录数目的长度
	 
	for(int i=1;i<=m;i++){    //阶乘 
		for(int j=0;j<number;j++){		 //数组须要操做有数的位置 
		    s = pass;	
			pass = (num[j]*i + pass)/10;  //肯定进位 
			num[j] = (num[j]*i + s)%10;	  //肯定后一位 
		} 
		s = pass;
       	while(pass > 0)                   //输出进位的位数 多是1位数 多是2位数，多是3位数 。。。。 
		{
            count++;
            pass = pass/10;            
		}
		while(count){
			num[number] = s%10;           //存储进位的值 
			number++;
			s = s/10;
			count--;
		}
        pass = 0;
	}
	return number;                       //返回长度 
}
int main(){
	int n;
	cin>>n;         //输入须要计算数目 
	while(n--){
		int m;
		cin>>m;
		cout<<get(m)<<endl;   //获得每一个样例结果 
	}
} 

后面发现对于N!来讲是比较快的了，可是对于n = 1000000时仍然会超时，因而乎决定不能这么直接

解题思路：

1.能够暴力，N的阶乖的位数等于LOG10（N！）＝LOG10（1）＋.....LOG10（N）；

2.Stirling公式：n!与√(2πn) * n^n * e^(-n)的值十分接近
故log10(n!) = log(n!) / log(10) = ( n*log(n) - n + 0.5*log(2*π*n))/log(n);

参考的是： http://blog.csdn.net/ultimater/article/details/7884951

 

因而乎比较简单

 

方法1：

#include<iostream>
#include<cmath>
using namespace std;

double get(int m){
    double cnt = 0;
	for(int i=2;i<=m;i++){
		cnt += log10(i);
	}
	return cnt;
}
int main(){
	int n;
	cin>>n;         //输入须要计算数目 
	while(n--){
		int m;
		cin>>m;
		cout<<1+(int)get(m)<<endl;   //获得每一个样例结果 
	}
} 

 

方法2：

#include<iostream>
#include<cmath>
using namespace std;
#define PI    3.141592653
#define ln10  log(10.0)

double get(int N){
    double cnt = 0;
    cnt = ceil((N*log(double(N))-N+0.5*log(2.0*N*PI))/ln10);
	return cnt;
}
int main(){
	int n;
	cin>>n;         //输入须要计算数目 
	while(n--){
		int m;
		cin>>m;
		cout<<(int)get(m)<<endl;   //获得每一个样例结果 
	}
} 

关于log10()能够参考 ： https://msdn.microsoft.com/zh-cn/library/t63833dz.aspx；

关于ceil()函数，向上取整 ，能够参考： https://baike.baidu.com/item/ceil/10931457?fr=aladdin

仍是数学公式好，死算的极可能GG