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
题解 ： 这道题本来不会，看了别人博客才会的；
log10(n)+1 -> 代表着 n 这个数的位数；
那么，n!的位数就等于 log10(n!)+1 ->log10(n)+log10(n-1)+..........+log10(2)+log10(1)+1;
#include<stdio.h>#include<math.h>int main(){ int t,n; scanf("%d",&t); while(t--){ scanf("%d",&n); double sum=0; for(int i=1;i<=n;i++){ sum+=log10(i); } sum+=1; printf("%d\n",(int)sum); } return 0;}
