Educational Codeforces Round 42 (Rated for Div. 2) C

C. Make a Square

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a positive integer 04 is incorrect).

In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.

Determine the minimum number of operations that you need to consistently apply to the given integer n to make from it the square of some positive integer or report that it is impossible.

An integer y.

Input

The first line contains a single integer 1≤n≤2⋅109). The number is given without leading zeroes.

Output

If it is impossible to make the square of some positive integer from -1. In the other case, print the minimal number of operations required to do it.

Examples

input

Copy

8314

output

Copy

2

input

Copy

625

output

Copy

0

input

Copy

333

output

Copy

-1

Note

In the first example we should delete from 9.

In the second example the given 25, so you should not delete anything.

In the third example it is impossible to make the square from -1.

题意：判断x是否是一个完全平方数，如果不是，最少能删去几个数让他是完全平方数，如果有，输出删去的最少操作次数，若没有，输出-1；

思路：暴力枚举，从1开始，到sqrt（x）位置，看是否满足这个数是x的子串，如果是，存下它的操作次数与min进行比较，输出 min；

小知识点：to_string 函数，可以将一串数字转化成字符串

#include<iostream>  
#include<string>  
  
using namespace std;  
  
int main()  
{  
    string s1 = "2018.11";  
    int a1 = stoi(s1);  
    double c = stod(s1);  
    float d = stof(s1);  
    int a2 = a1 + 1;  
    string b = to_string(a1);  
    b.append(" is a string");  
      
    cout << a1 <<"/"<<c<<"/"<<d<<"/"<<a2<<"/"<<b<< endl;  
}  

AC代码

#include <bits/stdc++.h>

#define x first
#define y second
#define pb push_back
#define mp make_pair
#define low_b lower_bound
#define up_b upper_bound
#define all(v) v.begin(), v.end()

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,int> pli;
typedef pair<ll,ll> pll;

inline void Kazakhstan(){
    ios_base :: sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
}

const int N = 2e9;

int main(){
    Kazakhstan();
    string s;

    cin >> s;
    int mn = INT_MAX;
    for(int i = 1; i * i <= N; i++){
        string t = to_string(i * i);
        int k = 0;
        bool w = 0;
        for(int j = 0; j < s.size(); j++){
            if(t[k] == s[j])k++;
            if(k == t.size()){ 
                w = 1;
                break;
            }
        }
        if(w){
            mn = min(mn, int(s.size() - t.size())); 
        }
    }
    if(mn == INT_MAX)cout << "-1";
    else cout << mn;
}

View Code