digital square DFS

Given an integer N,you should come up with the minimum nonnegative integer M.M meets the follow condition: M ^2%10 ^x=N (x=0,1,2,3…)

Input
The first line has an integer T( T< = 1000), the number of test cases.
For each case, each line contains one integer N(0<= N <=10 9), indicating the given number.

Output
For each case output the answer if it exists, otherwise print “None”.

Sample Input
3
3
21
25

Sample Output
None
11
5

需要找一个满足条件的最小的M使其等式成立，考得是DFS，需要从最后一位开始找，满足条件就入队，直到符合条件为止，不然输出None。代码参考网上大神的，侵删。

#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
struct Node{
	long long int num;
	long long int t;
	friend bool operator < (Node a,Node b)
    {
        return a.num>b.num;//用到优先队列来符合最小，需要重载运算符
    }
};
struct Node x,y;
void bfs(int n,int flag)
{
    priority_queue<Node>q;
	for(int i=0;i<10;i++)
	{
		if(i*i%10==n%10)
		{
			x.num=i;
			x.t=10;
			q.push(x);//寻找符合条件的个位数
		}
	}
	while(!q.empty())
	{
		x=q.top();
		if(x.num*x.num%flag==n)
		{
			cout<<x.num<<endl;
			return ;
		}
		q.pop();
		for(int i=0;i<10;i++)
		{
			y.num=x.num+i*x.t;
			y.t=x.t*10;
			if(y.num*y.num%y.t==n%y.t)
				q.push(y); //在所有符合条件的数中继续搜索符合条件就入队
		} 
	}
	cout<<"None"<<endl;
}
int main()
{
	int t,n,flag;
	cin>>t;
	while(t--)
	{
		cin>>n;
		flag=1;
		int temp=n;
		while(temp)
		{
			flag*=10;
			temp/=10;
		}
		bfs(n,flag);
	}
	return 0;
}