Educational Codeforces Round 42 C. Make a Square(字符串操作)

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最新推荐文章于 2023-03-13 16:03:25 发布

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本文介绍了一个算法问题：如何通过删除给定正整数中的某些数字，使其变为另一个正整数的平方。文章提供了完整的C++代码实现，并解释了其工作原理。

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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 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=y2 for some positive integer y

.
Input

The first line contains a single integer n
(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 n

, print -1. In the other case, print the minimal number of operations required to do it.
Examples
Input
Copy

8314

Output
Copy

Input
Copy

625

Output
Copy

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.

比赛的时候想着将数字看成字符串的,但是没有想清楚就写了,最后终测还是挂掉了.
看成字符串处理,可以不考虑前缀零.然后转换成数字判断是否满足条件即可

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rep1(i,b,a) for(int i=b;i>=a;i--)
using namespace std;
const int N=2e5+100;
ll arr[N];
vector<int> G;
struct node
{
    string x;
    int d;
};
int main()
{
    string str;
    cin>>str;
    node t={str,0};
    map<ll,int>mp;
    map<string,int>mp1;
    for(ll i=1;i<1e5;i++)
    {
        mp[i*i]=1;
    }
    queue<node>q;
    q.push(t);
    mp1[str]=1;
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        string s=t.x;
        int sum=(int)s[0]-'0';
        for(int i=1;i<s.size();i++)
        {
            sum=sum*10+int(s[i]-'0');
        }
        //cout<<sum<<endl;
        if(s[0]!='0'&&mp.count(sum))
        {
            cout<<t.d<<endl;
            return 0;
        }
        for(int i=0;i<s.size();i++)
        {
            string s1=s.substr(0,i)+s.substr(i+1,s.size()-1-i);
            if(!mp1.count(s1))
            {
                mp1[s1]=1;
                node tt={s1,t.d+1};
                q.push(tt);
            }
        }
    }
    cout<<-1<<endl;
    return 0;
}