**Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 50667 Accepted Submission(s): 24808 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 2 10 20
Sample Output 7 19**
思路：这是一个相当纯粹的数学题，首先我们知道n的位数一般等于log10(n)+1,所以n！的位数即为log10(n!)+1 = log10(1)+log10(2)+…+log10(n)+1. 代码：
#include <iostream> #include <cmath> using namespace std; int main(){ int T; cin>>T; while(T--){ int num; double n = 0; cin>>num; for(int i=1;i<=num;i++){ n+=log10(i); } cout<<(long int)n+1<<endl; } return 0; }