Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 40782    Accepted Submission(s): 19958

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

思路：

原来还有这种操作：求n的位数为（int）log10（n）+1，阶乘为相加后加1

code

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #define INF 0x3f3f3f3f#define N 85using namespace std;int main(){ int t,n,i; double sum; scanf("%d",&t); while(t--){ scanf("%d",&n); sum=0; for(i=1;i<=n;i++){ sum+=log10(i); } printf("%d\n",(int)sum+1); } return 0;}