HDOJ 1018 Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18071    Accepted Submission(s): 8115

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)

 

Recommend

JGShining

 

裸算必然TLE。。。我无耻的翻解题报告了。 数学题

算法一：利用公式 log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)

#include <iostream>#include <cmath> using namespace std;int main(int argc, char *argv[]){int t,i,j,n; double sum;cin>>t;for (i=0;i<t;++i){cin>>n; sum=0.0;for(j=1;j<=n;++j)sum+=log10(j);n=round(sum+0.5);cout<<n<<endl;}return 0;}

差点TLE。

算法二：

《计算机程序设计艺术》中给出了另一个公式
   n! = sqrt(2*π*n) * ((n/e)^n) * (1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3))
   其中 π = acos(-1)    e = exp(1)

有了这个公式就好好办了，两边对10取对数：
忽略log10(1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) ≈ log10(1) = 0
得到公式  log10(n!) = log10(sqrt(2 * pi * n)) + n * log10(n / e)
用C++写时round（n+0.5）会出错，要用（int）n+1。。。。。C的取整函数真纠结啊，怀念Pascal中

果断　0MS ＡＣ

#include <iostream>#include <cmath> using namespace std;int main(int argc, char *argv[]){int t,n; double PI = acos(-1.0),e = exp(1.0); cin>>t; for (int i=0;i<t;++i) {cin>>n; n=log10(sqrt(2*PI*n))+(n*log10(n/e))+1; cout<<int(n)<<endl;  }return 0;}

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