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 $\mathrm{nn,\; written\; without\; leading\; zeroes\; (for\; example,\; the\; number 04 is\; incorrect).}$

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 $\mathrm{nn to\; make\; from\; it\; the\; square\; of\; some\; positive\; integer\; or\; report\; that\; it\; is\; impossible.}$

An integer $\mathrm{xx is\; the\; square\; of\; some\; positive\; integer\; if\; and\; only\; if x=y2x=y2 for\; some\; positive\; integer yy.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 2\cdot 1091\le n\le 2\cdot 109).\; The\; number\; is\; given\; without\; leading\; zeroes.}$

Output

If it is impossible to make the square of some positive integer from $\mathrm{nn,\; print -1.\; In\; the\; other\; case,\; print\; the\; minimal\; number\; of\; operations\; required\; to\; do\; it.}$

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 $\mathrm{83148314 the\; digits 33 and 44.\; After\; that 83148314 become\; equals\; to 8181,\; which\; is\; the\; square\; of\; the\; integer 99.}$

In the second example the given $\mathrm{625625 is\; the\; square\; of\; the\; integer 2525,\; so\; you\; should\; not\; delete\; anything.}$

In the third example it is impossible to make the square from $\mathrm{333333,\; so\; the\; answer\; is -1.}$

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;
}