hdu 1018 Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24081    Accepted Submission(s): 10926

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

 

Source

Asia 2002, Dhaka (Bengal)

 

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输入一个n，求n阶乘结果有多少位

室友胜哥考我的一道题，我都没怎么想就百度了。。。

这个臭毛病一定要改，遇到题多思考，不要上手就百度

有两种方法

1、用log的性质log(n!) = log1 + log2 + ... + logn; log(n!)即n!的位数

2、用斯特林公式;

当然斯特林公式也要用到log思想来求位数

需要注意的是log的使用，大概hdu上的c++版本不够，我的编译器可以通过，但是交题后显示CE

改正后的代码如下：

#include <map>#include <cmath>#include <vector>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define esp 1e-9#define MAXN 10010#define ll long long#define INF 0x7FFFFFFF#define BUG system("pause")#define SW(a,b) a^=b;b^=a;a^=b;#define Pi 3.1415926using namespace std;int main(void){int n;while(cin >> n){while(n--){int a;cin >> a;double sum = 0;for(int i=1; i<=a; ++i){sum += (log(i*1.0)/log(10.0));} //cout << (int)sum+1 << endl;cout << (int)(0.5*(log(2.0*Pi*a)/log(10.0))+a*1.0*(log(a*1.0)-1.0)/log(10.0))+1 << endl;}} return 0;}