HDU 1018 Big Number（斯特林公式 或 暴力）

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39915    Accepted Submission(s): 19434

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

 

1.斯特林公式:N！=sqrt(2*pi*n)*pow((n/e),n);

N！的位数=log10(2*pi*n)/2+n*log10(n/e)+1;

#include <stdio.h>#include<math.h>double pi=3.14159265358;double e=2.71828182;int main(int argc, char *argv[]){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int ans=log10(2*pi*n)/2+n*log10(n/e)+1;printf("%d\n",ans);}return 0;}

2.暴力:每次超过10就开始去掉当前的数位 注意！！！！存阶乘的sum要用 double用long long会wa

#include <stdio.h>int main(int argc, char *argv[]){int t;scanf("%d",&t);while(t--){long long n;scanf("%lld",&n);double sum=1;int i;int cnt=1;for(i=2;i<=n;i++){sum=sum*i;if(sum>=10){while(sum>=10){sum/=10;cnt++;}}}printf("%d\n",cnt);}return 0;}