Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29336 Accepted Submission(s): 13454
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
Sample Output
ps:公式題
題意:一個數n的階層的位數等於1+log10(i)+.....+log10(n)
#include<cmath>
#include<cstdio>
using namespace std;
int main()
{
int num,t;
scanf("%d",&t);
while(t--)
{
double a=1.0;
scanf("%d",&num);
for(int i=1;i<=num;i++)
a+=log10(i);
printf("%d\n",(int)a);
}
return 0;
}