Digital Square

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1827    Accepted Submission(s):
714

Problem DescriptionGiven an integer N,you should come up with the minimum
nonnegative integer M.M meets the follow condition:
M²%10^x=N (x=0,1,2,3….) InputThe first line has an integer T( T< = 1000), the
number of test cases.
For each case, each line contains one integer N(0<=
N <=10⁹), indicating the given number. OutputFor each case output the answer if it exists, otherwise
print “None”.  Sample Input332125 Sample OutputNone115 Source2012
Multi-University Training Contest 10 Recommendzhuyuanchen520   |   We have carefully selected several
similar problems for you:  4390 4398 4397 4396 4395  搜索题，首先证明出N位后缀只与M的后N位有关。比如三位数100a+10b+c平方后展开为 10000a^2+2000ab+100b^2+200ac+20bc+c^2很显然，平方后的最后一位只与c有关最后两位只与bc有关，最后三位abc都有关。那我们只需要BFS一下，不断地找满足最后指定位数的数，1位，2位，……直到找到第一个满足条件的。 题意：给出n，求出最小的m，满足m^2  % 10^k = n，其中k=0,1,2 附上代码： 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <queue>
 using namespace std;
 __int64 n;
 struct node
 {
     __int64 a;
     int x;
     friend bool operator < (node n1,node n2)  //优先队列，必须输出最小的数
     {
         return n1.a>n2.a;
     }
 } s1,s2; void BFS()
 {
     priority_queue <node> q;
     while(!q.empty())
         q.pop();
     __int64 t;
     s1.a=;
     s1.x=;
     q.push(s1);
     while(!q.empty())
     {
         s1=q.top();
         q.pop();
         t=(__int64)pow(10.0,s1.x);  //从低位向高位搜
         if(s1.a*s1.a%t==n)      //找到了就直接输出
         {
             printf("%I64d\n",s1.a);
             return;
         }
         for(int i=; i<; i++)
         {
             s2.x=s1.x+;     //每次都向前进一位
             s2.a=s1.a+i*t;
             if(s2.a*s2.a%(t*)==n%(t*))  //一位一位的匹配，匹配成功后，则放入队列
                 q.push(s2);
         }
     }
     printf("None\n");
 }
 int main()
 {
     int T,m,i,j;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         if(n%==||n%==||n%==||n%==)  //根据乘法的规律，任何一个数平方后的个位数不可能是这些数
         {
             printf("None\n");
             continue;
         }
         BFS();
     }
     return ;
 }