Algorithm

Problem link- https://www.spoj.com/problems/LENGFACT/

LENGFACT - Factorial length

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Given integer n, print length of n! (which is factorial of n).

Input

The first line of the standard input contains one integer t (t < 10001) which is the number of test cases.

In each of the next t lines there is number n (0 <= n <= 5*10^9).

Output

For each test, print the length of n! (which is factorial of n).

Example

Input:
3
1
10
100

Output:
1
7
158

 

Code Examples

#1 Code Example with Python Programming

Code - Python Programming

import math

T = int(input())
for _ in range(T):
    N = int(input())
    if N < 2:
        print(1)
    else:
        print(math.floor((math.log(2 * math.pi * N) / 2 + \
                     N * (math.log(N) - 1)) / math.log(10) \
                     ) + 1)
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Input

x
–
+
cmd
3
1
10
100

Output

x
–
+
cmd
1
7
158

#2 Code Example with C++ Programming

Code - C++ Programming

#include<iostream7gt;
#include<cmath>
using namespace std;   
int main()
{
int t;
cin>>t;
double n;
double pi = 2*acos(0);
long long int len,l;

while(t-->0)
{
        cin>>n;
        if(n==0.00||n==1.00)l=1;

        else{
        l=(((log(2.00*pi*n)/2.00)+(n*(log(n)-1.00)))/log(10.0))+1;
        }
       
        
        cout<<l<<endl;
}
return 0;
}
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Input

x
–
+
cmd
3
1
10
100

Output

x
–
+
cmd
1
7
158
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Demonstration

SPOJ Solution-Factorial length-Solution in C, C++, Java, Python

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