Problem:
Score : 300 points

Problem Statement

You are given an integer N.

For two positive integers A and B, we will define F(A,B) as the larger of the following: the number of digits in the decimal notation of A, and the number of digits in the decimal notation of B.

For example, F(3,11) = 2 since 3 has one digit and 11 has two digits.

Find the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Constraints

1 \leq N \leq 10^{10}

N is an integer.

Input

The input is given from Standard Input in the following format:

N

Output

Print the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Sample Input 1

10000

Sample Output 1

3

F(A,B) has a minimum value of 3 at (A,B)=(100,100).

Sample Input 2

1000003

Sample Output 2

7

There are two pairs (A,B) that satisfy the condition: (1,1000003) and (1000003,1). For these pairs, F(1,1000003)=F(1000003,1)=7.

Sample Input 3

9876543210

Sample Output 3

6

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.*;
import java.math.*;

 class Main {
     static boolean[][] graph = new boolean[8][8];

     public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long N = sc.nextLong();
        long Nw= N;
        long[] yaku=new long[100000];
        int k=0;
        long root = (long) Math.pow(N,1.0/2.0)+1;
        
        for(long i=1;i<root;i++) {
        	if(Nw%i==0) {
        		yaku[k]=i;
//                System.out.println(i);
        		k++;
        	}
        }
//        System.out.println(Nw);
//        System.out.println("k="+k);
        long ans=999;
        for(int i=k-1;i>=0;i--) {
            long A=yaku[i];
            long B=N/yaku[i];
            long result=Math.max(A, B);
//          System.out.println("A=" + A);
//          System.out.println("B=" + B);
            if (String.valueOf(result).length()<ans)ans=String.valueOf(result).length();
        }
        System.out.println(ans);
     }   
    static int dfs(int first, int max, boolean[] visited) {
    	int ret = 0;
    	boolean all_visited = true;
    	for(int i=0;i<max;i++) {
    		if(visited[i]==false) {
    			all_visited=false;    			
    		}
    	}
    	if (all_visited) {
    		return 1;
    	}
    	
    	for(int i=0;i<max;i++) {
    		if(graph[first][i]==false) {
    			continue;
    		}
    		if(visited[i]==true) {
    			continue;
    		}
    		visited[i]=true;
			ret += dfs(i,max,visited);
			visited[i] = false;
    	}
    	return ret;
    }
}