C - Digits in Multiplication (atcoder.jp)

给定一个数n，将n拆成 n=a * b , 定义F(A,B) = A,B之中 位数的最大值

求这个F(A,B) ,满足A * B  = n

#include <iostream>
#include <vector>
#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <set>
#include <stack>
#include <algorithm>
#define x first
#define y second
#define pb emplace_back
#define fu(i,a,b) for(int i=a;i<=b; ++ i)
#define fd(i,a,b) for(int i=a;i>=b;	-- i)
#define endl '
'
#define ms(x,y) memset(x,y,sizeof x)
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef vector<vector<LL>> VVL;
typedef vector<vector<int>> VVI;
typedef vector<LL> VL;
typedef vector<bool> VB;
typedef vector<int> VI;
typedef vector<string> VS;
typedef pair<int,int> PII;
typedef vector<PII> VPII;
typedef pair<PII,int> PIII;
typedef pair<double,double> PDD;
typedef pair<double,int> PDI;
typedef pair<char,int> PCI;
typedef pair<string,int> PSI;
typedef pair<int,string> PIS;
typedef pair<int,char> PIC;
typedef pair<LL,LL> PLL;
typedef __int128 i128;
typedef unsigned long long ULL;
const int N =100+ 10 ,M = N * 2 ,INF = 0x3f3f3f3f,P = 131;
const double eps = 1e-8,DNF = 1e18;
const int mod = 1e9 +  7 ,base= 20010;
const LL LNF=(LL) INF * INF;


void solve()
{
	LL n;cin >> n;	
	
	int ans= to_string(n).size();	
	
	fu(i,1,n/i)
	{
		if(n % i ) continue;
		LL t = n / i;
 		int cnt = max(to_string(i).size(),to_string(t).size());
 		ans = min(ans,cnt)	;
	}
	cout << ans << endl;
	
					
}

signed main()
{
//  freopen("1.txt","r",stdin);
//	#define int long long 
//	init(N-1);
    ios
//	cout << fixed<<setprecision(2);
    int t=1;
//   	cin>>t;
    int now = 1;
    while(t -- )
    {
//      cout<<"Case "; 
//      cout<<"Case #"; 
//      cout<<"Scenario #"; 
//      cout<< now ++ <<": ";
//      cout<< now ++ <<": 
";
		solve();
    }


    return 0;
}