T^T（2）

Description

T ^ T This is much like a cry of expression is not it! In fact, it is the power of friends ~ T-T.
When T is relatively large when T ^ T will be very large, and now this number if you find a total of how many bits on it.

Input

Input comprises a plurality of sets of test data, the test data is only one number for each T (0 <T <2 ^ {31}) T (0 <T <2 (31)).

Output

Please output T ^ T is a total number of digits.

Sample Input 1

3
5

Sample Output 1

2
4

Problem-solving ideas

A thought:

Brute force.
Direct calculation of n n-th power, and then converted to a string, then the count
, of course, this method certainly bother

Thinking two:

Because we did a quick question of power. Stupid I wanted to try
the following code:

#include<stdio.h>
#include<math.h>
long long int f(int a,int b)
{
    int r=1;
    while(b!=0)
    {
        if(b%2==1)
        {
            r*=a;
        }
        a*=a;
        b>>=1;
    }
    return r;
}
int main()
{
    int n;
    long long int f(int a,int b);
    while(scanf("%d",&n)!=EOF)
    {
        printf("%.0f\n",floor(log10(f(n,n)))+1);
    }
    return 0;
}

The results do not guess ah! time out!
So, I am such a waste of time
to continue to observe. You see this log function, it is not there-something? ? ?
log (a ^ b) = b
* log (a) remember him! very useful!
ok

Thinking three:

such. Have a problem. The extra log is doing it
seek digits. log10 = 1; log100 = 2
then the number of bits required number n can be
n * log (n) +1
code is as follows:

#include<stdio.h>
#include<math.h>
int main()
{
    long long int n,m;
    while(scanf("%lld",&n)!=EOF)
    {
        m=floor(n*log10(n))+1;
        printf("%lld\n",m);
    }
    return 0;
}

Weeping, Method two rows so many really grueling.

11.26（T^T）