codeforces 962 C. Make a Square

最新推荐文章于 2024-07-29 22:43:34 发布

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最新推荐文章于 2024-07-29 22:43:34 发布

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本文介绍了一个算法问题，即如何通过删除数字将一个正整数变为完全平方数，并给出了解决方案。文章详细解释了输入输出格式及示例，提供了一段C++代码实现。

You are given a positive integer $n$ , 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 $n$ to make from it the square of some positive integer or report that it is impossible.

An integer $x$ is the square of some positive integer if and only if $x = y^{2}$ for some positive integer $y$ .

Input

The first line contains a single integer $n$ ( $1 \leq n \leq 2 \cdot 10^{9}$ ). The number is given without leading zeroes.

Output

If it is impossible to make the square of some positive integer from $n$ , 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 $8314$ the digits $3$ and $4$ . After that $8314$ become equals to $81$ , which is the square of the integer $9$ .

In the second example the given $625$ is the square of the integer $25$ , so you should not delete anything.

In the third example it is impossible to make the square from $333$ , so the answer is -1.

这个最多只有10位。。。。

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int maxn = 1e5+7;
ll INF = 1e8+7, t;

string s;
vector<ll> a;


bool solve()
{
    if(a.size() == 0) return false;
    else if(a[0] == 0) return false;
    else {
        ll res = 0;
        for(int i = 0; i < a.size(); i++)
        {
            res = res*10+a[i];
        }
        ll sres = sqrt(res);
        if(res == sres*sres) return true;
        else return false;
    }
}

int main()
{
    cin >> s;
    int n = s.length();
    ll len = 1<<n, res = INF;
    for(int i = 0; i <= len; i++)
    {
        a.clear();
        for(int j = 0; j < n; j++)
        {
            if(i>>j&1) a.push_back(s[j] - '0');
        }
        if(solve()) res = min(res, (ll)(n - a.size()));
    }
    if(res == INF) res = -1;
    printf("%d\n",res);
}